Probability

Given: A coin is tossed once.

To Find: Write its sample space?

Explanation: Here, the coin is tossed only once,

Then, there are two probability either Head(H) or Tail(T)

So, Sample will be

S = {H, T}

Where, H denotes Head and T denotes Tail

Hence, The sample is {H, T}

Given: If Coin is tossed twice times.

To Find: Write the sample space associated to this experiment.

Explanation: Here, two coins are tossed, that means two probability will occur at same time

So, The sample space will be

S={HT, TH, HH, TT}

Hence, Sample space is {HT, HH, TT, TH}

Given: If a coin is tossed three times .

To Find: Write the sample space for the given experiment.

Explanation: Here, the coins is tossed three time, then the no. of samples

2³=8

So, The sample space will be

S={HHH, TTT, HHT, HTH, THH, HTT, THT, TTH}

Hence, The sample space is {HHH, TTT, HHT, HTH, THH, HTT, THT, TTH}

Given: A coin is tossed four times.

To Find: Write the sample space for the given experiment.

Explanation: Here, The coins is tossed four time, then the no. of samples

2⁴=16

So, The sample space will be

S={HHHH, TTTT, HHHT, HHTH, HTHH, THHH, HHTT, HTTH, HTHT, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH}

Hence, The sample space is {HHHH, TTTT, HHHT, HHTH, HTHH, THHH, HHTT, HTTH, HTHT, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH}

Given: Two dice are thrown.

To Find: Write the sample space for the given experiment.

Explanation: We know there are 6 faces on a dice. Contains (1, 2, 3, 4, 5, 6).

But, Here two dice are thrown, then we have two faces of dice (one of each)

So, The total sample space will be 6² = 36

Now, the sample space is:

S={(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1)(3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), 5, 4), (5, 6), (5, 5), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Given: Three dice is rolled together.

To Find: What is the total number of elementary events.

Explanation: Here, three dice are thrown together,

And, There are 6 faces on die,

So, The total number of elementary event on throwing three dice are

6×6×6=216

Hence, The total number is 216

Given: A coin is tossed and a die is thrown.

To Find: Write the sample space for the given experiment.

Explanation: Here, The coin is tossed and die is thrown.

We know, when coin is tossed there will be 2 events either Head or Tail.

And, when die is thrown then there will be 6 faces (1, 2, 3, 4, 5, 6)

SO, The total number of Sample space together is 2×6 = 12

S={(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}

Hence, Sample space are {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}

Given: A coin is tossed and the a die is rolled.

To Find: Write the sample space for the given experiment.

Explanation: Here, we have a coin and a die,

We know, when coin is tossed there will be 2 event Head and tail,

According to question, If Head occurs on coin then Die will rolled out otherwise not.

So, the sample spaces are :

S={(T, (H, 1)(H, 2), (H, 3), (H, 4), (H, 5), (H, 6)}

Hence, Sample space is (T, (H, 1)(H, 2), (H, 3), (H, 4), (H, 5), (H, 6)

Given: A coin is tossed twice. If the second throw results I a tail, A die is thrown.

To Find: Write the sample space for the given experiment.

Explanation: When a coin tossed twice, Then sample spaces for only coin will be: {HH, TT, HT , TH}

Now, According to question , when we get Tail in second throw, then a dice is thrown.

So, The total number of elementary events are 2+(2×6)=14

And sample space will be

S={HH, TH, (HT, 1), (HT, 2), (HT, 3), (HT, 4), (HT, 5), (HT, 6), (TT, 1), (TT, 2), (TT, 3), (TT, 4), (TT, 5), (TT, 6)}

Hence, this is the sample space for given experiment.

Given: A coin is tossed and a die is rolled.

To Find: Write the sample space for the given experiment.

Explanation: In the given experiment, coin is tossed and if the outcome is tail then, die will be rolled.

The possible outcome for coin is 2 = {H, T}

And, The possible outcome for die is 6 = {1, 2, 3, 4, 5, 6}

If the outcome for the coin is tail then sample space is S1={(T, 1)(T, 2)(T, 3)(T, 4)(T, 5)(T, 6)}

If the outcome is head then the sample space is S2={(H, H)(H, T)}

So, The required outcome sample space is S=S1S2

S={(T, 1)(T, 2)(T, 3)(T, 4)(T, 5)(T, 6)(H, H)(H, T)}

Hence, The sample space for the given experiment.

Given: A coin is tossed and there is box which contain 2 red and 3 black balls.

To Find: Write the sample space for the given experiment.

Explanation: when coin is tossed , there are 2 outcome {H, T}

According to question, If tail turned up, the a ball is drawn from a box.

So, Sample for This experiment S₁ = {(T, R₁)(T, R₂)(T, B₁)(T, B₂)(T, B₃)}

Now, If Head is turned up, then die is rolled

So, Sample space for this experiment S₂={(H, 1)(H, 2)(H, 3)(H, 4)(H, 5)(H, 6)}

The required sample space will be S = S₁ S₂

So, S={(T, R₁), (T, R₂), (T, B₁), (T, B₂), (T, B₃), (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)}

Hence, S is the elementary events associated with the given experiment.

Given: A coin is tossed repeatedly until comes up fro the first time.

To Find: Write the sample space for the given experiment.

Explanation: In the given Experiment, a coin is tossed and if the outcome is tail the experiment is over ,

And, if the outcome is Head then the coin is tossed again.

In the second toss also if the outcome is tail then experiment is over, otherwise coin is tossed again.

This process continues indefinitely

SO, The sample space for this experiment is

S={T, HT, HHT, HHHT, HHHHT…}

Hence, S is th sample space for the given experiment.

Given: A box contains 1 red and 3 black balls.

To Find: Write the sample space for the given experiment.

Explanation: We have 1 red and 3 black balls in a box.

Let Assume Red = R

Let Assume Blue = B

Since, two balls are drawn at random without replacement,

So, The sample spaces for this experiment is:

S={(R, B₁), (R, B₂), (R, B₃)(B₁, B₂)(B₁, B₃)(B₁, R)(B₂, R)(B₂, B₁)(B₂, B₃)(B₃, R)(B₃, B₁)(B₃, B₂)}

Hence, S is the sample spaces for given experiment.

Given: A pair of dice is rolled , a coin is tossed.

To Find: Write the sample space for the given experiment.

Explanation: A pair of dice is rolled,

Then, No. of elementary events are 6²=36

Now, If outcomes is doublet means (1, 1)(2, 2)(3, 3)(4, 4)(5, 5)(6, 6), then a coin is tossed.

If coin is tossed then no. of sample spaces is 2

So, The total no. of elementary events including doublet = 6×2=12

Thus, The Total number of elementary events are 30+12 =42

Hence, 42 events will occur for this experiments.

Given: There is a coin which tossed twice.

To Find: Write the sample space for the given experiment.

Explanation: A coin is tossed twice , So the outcomes are

S₁={HH, HT, TH, TT}

Now, If the second drawn result is head , the a die is rolled then the elementary events is

S₂={(HH, 1), (HH, 2)(HH, 3), (HH, 4)(HH, 5), (HH, 6), (HH, 1), (TH, 2)(TH, 3), (TH, 4)(TH, 5), (TH, 6)}

Thus, The total sample space for the experiment is S=S₁ S₂

S={(HH), (HT), (TH), (TT)(HH, 1), (HH, 2)(HH, 3), (HH, 4)(HH, 5), (HH, 6), (HH, 1), (TH, 2)(TH, 3), (TH, 4)(TH, 5), (TH, 6)}

Hence, S is the sample space for given experiment.

Given: A bag contains 4 identical red balls and 3 identical black balls.

To Find: Write the sample space for the given experiment.

Explanation: A bag contains 4 red balls and 3 black balls.

Let us Assume Red = R

Let us Assume Black = B

Now, A ball is drawn in first attempt, So elementary events is

S₁={R, B}

And, The bag will put into the bag and draw again, then elementary events are

S₂={R, B}

Thus, The total sample space associated is S=S₁S₂

S={RR, RB, BR, BB}

Hence, S is the sample space for given experiment.

Given: There are three sampling item.

To Find: Write the sample space for the given experiment.

Explanation: In the random sampling , three items are selected so it could be

(a) All defective (D)

(b) All non-defective(N)

(c) Combination of both defective and non defective

We have 2 category (N and D) and we have three condition

So, The number of sample is 2³ = 8

Sample space associated with this experiment is

S={DDD, NDN, DND, DNN, NDD, DDN, NND, NNN)

Hence, S is the sample space for given experiment

(i) To Find: Write the sample space for the given experiment.

Explanation: Here, The family has 2 children

Let us Assume Boy = B

Let us Assume Girl = G

So, The number of sample spaces for 2 children is 2²=4

Sample space are, S={(B₁, B₂), (G₁, G₂), (G₁, B₂), (B₁, G₂)}

Where , number 1 and 2 are represent elder and younger.

Hence, S is the sample space for given experiment.

(ii) Explanation: Here, The family has 2 children, So the possibility of boys in a family is:

(a) No boys only girl

(b) One boy and one girl

(c) Two boys only

So, The Sample space for the given condition is:

S={0, 1, 2}

Hence, S is the sample spaces for the given experiment.

Given: There are three colored dice of red , white and black color. These dice are placed in bag.

To Find: Write the sample space for the given experiment.

Explanation: A dice has 6 faces containing numbers (1, 2, 3, 4, 5, 6)

Let us Assume Red = R

Let us Assume White = W

Let us Assume Black = B

According to question, when Dice is selected , then rolled.

Firstly, selected Red Dice then, Possible samples are:

S_R={(R, 1), (R, 2)(R, 3), (R, 4), (R, 5), (R, 6)}

Then, White Dice will be selected , So the sample spaces are:

S_w={(W, 1), (W, 2), (W, 3), (W, 4), (W, 5), (W, 6)}

Lastly, Black Dice will be selected and rolled , So the sample spaces for Black die

S_B= {(B, 1)(B, 2)(B, 3)(B, 4)(B, 5), (B, 6)}

Thus, The total sample for the given experiment is

S=S_R S_W S_B

Hence, B is the sample space for the given experiment.

Given: 2 boys and 2 girls are in room P and 1 boy 3 girls are in room Q.

To Find: Write the sample space for the given experiment.

Explanation: Let us denote two boys and girls in room P as B₁ , B₂ and G_{1 ,} G₂ respectively

Let us denote 1 boy and 3 girls in room Q as B₃ and G₃, G₄, G₅ respectively.

According to the question,

Sample spaces for room S_p are

S_p={(P, B₁), (P, B₂), (P, G₁), (P, G₂)}

And, Sample spaces for room S_q are

S_q={(Q, B₃), (Q, G₁), (Q, G₂), (Q, G₃)}

Now, The total sample spaces for the given experiment:

S=

S={(P, B₁), (P, B₂), (P, G₁), (P, G₂), (Q, B₃), (Q, G₁), (Q, G₂), (Q, G₃)}

Hence, S is the sample space for the given experiment

Given: A bag contains one white and one red ball.

To Find: Write the sample space for the given experiment.

Explanation: Here, There are 1 white ball , 1 red ball in a bag

Let us denote White ball with W and Red Ball with W

According to question, when one ball is drawn, it may be a white or Red.

So, The sample space of drawing on white ball with replacement

S_w={(W, W), (W, R)}

Again, If red ball is drawn , a dice is rolled

So, The sample space for red ball with dice

S_R={(R, 1), (R, 2), (R, 3), (R, 4), (R, 5), (R, 6)}

Thus, The required sample space for the experiment is:

S=S_W S_R

So, S= {(W, W), (W, R), (R, 1), (R, 2), (R, 3), (R, 4), (R, 5), (R, 6)}

Hence, This is the sample space for given experiment

Given: A box contain1 white and 3 identical black balls.

To Find: Write the sample space for the given experiment.

Explanation: Here, only 1 white ball and 3 Black identical Balls.

Let us denote the white ball with W and Black ball with B

When two balls are drawn at random in succession without replacement,

So, The sample space for this experiment is

S={WB, BW, BB}

Given: A dice is rolled and then a coin is tossed once.

To Find: Write the sample space for the given experiment.

Explanation: A dice has 6 faces that are numbered from 1 to 6.

And, we know, 2, 4, 6 are even numbers and 1, 3, 5 are odd numbers.

Now, A coin has two face, Head(H) and Tail(T)

So, we can find the total sample spaces by combining both experiments as

S={(2, H), (2, T), (4, H), (4, T), (6, H), (6, T), (1, HH), (1, HT), (1, TH), (1, TT), (3, HH), (3, HT) , (3, TH), (3, TT), (5, HH), (5, HT), (5, TH), (5, TT)}

Hence, S is the required sample space for given experiment.

Given: A die is thrown repeatedly until a six comes up.

To Find: Sample space for the given experiment

Explanation: Here, The die is thrown repeatedly until a six comes up. It means

Since, 6 may come upon the first throw , second throw , third throw and so on until is obtained.

So, This process will continues indefinitely, then the sample space will be:

S={6, (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (1, 1, 6), (1, 2, 6), (1, 3, 6), (1, 4, 6), (1, 5, 6), (2, 1, 6), (2, 2, 6), (2, 3, 6), (2, 4, 6), (2, 5, 6)……..}

Hence, S is the sample space for given experiment.

Given: A coin is tossed.

To Find: Find the total number of elementary events and a total number of events associated with the random experiment.

Explanation: When a coin is tossed, there will be two possible outcomes, Head(H) and Tail(T).

Since, the no. of elementary events is 2 {H, T}

But, we know, if there are n elements in a set, then the number of total element in its subset is 2n.

Therefore, the total number of the experiment is 4,

So, there are 4 subset of S = {H}, {T}, {H, T} and Փ

Hence, there are 4 total events in a given experiment.

Given: Two coins are tossed once.

To find: How many events are elementary events

Explanation: We know, when Two coins are tossed then the no. of possible outcomes are 2² = 4

So, The Sample spaces are {HH, HT, TT, TH}

Hence, there are total 4 events associated with the given experiment.

Given: There are three coins tossed once.

To Find: Describe the events according to the subparts?

Explanation: when three coins are tossed, then the sample spaces are:

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

According to the question,

A = {HHH}

B = {HHT, HTH, THH}

C = {TTT}

D = {HHH, HHT, HTH, HTT}

Now,

Since, If the intersection of two sets are null or empty it means both the sets are Mutually Exclusive.

(i) Events A and B, Events A and C, Events B and C and events C and D are mutually exclusive.

(ii) Here, We know, If an event has only one sample point of a sample space, then it is called elementary events.

So, A and C are elementary events.

(iii) If There is an event that has more than one sample point of a sample space, it is called a compound event,

Since,

So, B and D are compound events.

Given: A dice is thrown once.

To Find: Find the given events, and also find the Also, find A ∪ B, A ∩ B, B ∩ C, E ∩ F, D ∩ F and

Explanation: In a single throw of a die, the possible events are:

S = {1, 2, 3, 4, 5, 6}

Now, According to the subparts of the question, we have certain events as:

(i) A = getting a number below 7

So, The sample spaces for A are:

A = {1, 2, 3, 4, 5, 6}

(ii) B = Getting a number greater than 7

So, the sample spaces for B are:

B = {Փ}

(iii) C = Getting multiple of 3

So, The Sample space of C is

C = {3, 6}

(iv) D = Getting a number less than 4

So, The sample space for D is

D = {1, 2, 3}

(v) E = Getting an even number greater than 4.

The sample space for E is

E = {6}

(vi) F = Getting a number not less than 3.

The sapmle space for F is

F = {3, 4, 5, 6}

Now,

A = {1, 2, 3, 4, 5, 6} and B = {Փ}

A = {1, 2, 3, 4, 5, 6} and B = {Փ}

B = {Փ} and C = {3, 6}

F = {3, 4, 5, 6}and E = {6}

E = {6} and D = {1, 2, 3}

And, For

S = {1, 2, 3, 4, 5, 6} and F = {3, 4, 5, 6}

Hence, These are the events for given ecperiment

When three coins are tossed, then the sample space is

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Now, The subparts are:

(i) The two events which are mutually exclusive are when,

A: getting no tails

B: getting no heads

Then, A = {HHH} and B = {TTT}

SO, The intersection of this set will be null.

Or, The sets are disjoint.

(ii) Three events which are mutually exclusive and exhaustive are:

A: getting no heads

B: getting exactly one head

C:getting at least two head

So, A = {TTT} B = {TTH, THT, HTT} and C = {HHH, HHT, HTH, THH}

Since,

(iii) The two events that are not mutually exclusive are:

A:getting three heads

B:getting at least 2 heads

So, A = {HHH} B = {HHH, HHT, HTH, THH}

Since

(iv) The two events which are mutually exclusive but not exhaustive are:

A:getting exactly one head

B: getting exactly one tail

So, A = {HTT, THT, TTH} and B = {HHT, HTH, THH}

It is because

Given: A dice is thrown twice. And each time number appearing on it is recorded.

To Find: Describe the given events.

Explanation: when the dice is thrown twice then the number of sample spaces are 6² = 36

Now,

The possibility both odd numbers are:

A = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}

Since, Possibility of both even numbers are:

B = {(2, 2)(2, 4)(2, 6)(4, 2)(4, 4)(4, 6)(6, 2)(6, 4)(6, 6)}

And, Possible outcome of sum of the numbers is less than 6

C = {(1, 1)(1, 2)(1, 3)(1, 4)(2, 1)(2, 2)(2, 3)(3, 1)(3, 2)(4, 1)}

Therefore,

(AՍB) = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5) (2, 2)(2, 4)(2, 6)(4, 2)(4, 4)(4, 6)(6, 2)(6, 4)(6, 6)}

(AՌB) = {Փ}

(AUC) = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5) (1, 2)(1, 4)(2, 1)(2, 2)(2, 3)(3, 1)(3, 2)(4, 1)}

(AՌC) = {(1, 1)(1, 3)(3, 1)}

Hence, (AՌB) = Փ and (AՌC)≠Փ, A and B are mutually exclusive, but A and C are not.

Given: Two dice are thrown.

To Find: Describe the given events.

Explanation: when two dice are thrown then the no. of possible outcomes are 6² = 36

Now, According to the question,

A = Getting an even number of the first die.

A = {(2, 1)(2, 2)(2, 3)(2, 4)(2, 5)(2, 6)(4, 1)(4, 2)(4, 3)(4, 4)(6, 1)(6, 2)(6, 3)(6, 4)(6, 5)(6, 6)}

B = Getting an odd number on the first dice

B = {(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5)(5, 6)}

C = Getting at most 5 as the sum of numbers on the two dices.

C = {(1, 1)(1, 2)(1, 3)(1, 4)(2, 1)(2, 2)(2, 3)(3, 1)(3, 2)(4, 1)}

D = Getting a sum greater than 5 but less than 10

D = {(1, 5)(1, 6) (2, 4)(2, 5) (2, 6)(3, 3)(3, 4)(3, 5)(3, 6)(4, 2)(4, 3)(4, 4)(4, 5)(5, 1) (5, 2)(5, 3) (5, 4)(6, 1) (6, 2)(6, 3)}

E = Getting at least 10 as the sum of numbers on the dices

{(4, 6)(5, 5)(5, 6)(6, 4)(6, 5)(6, 6)}

F = Getting an odd number on one of the dices

{(1, 2)(1, 4)(1, 6)(2, 1)(2, 3)(2, 5)(3, 2)(3, 4)(3, 6)(4, 1)(4, 3)(4, 5)(5, 2)(5, 4)(5, 6)(6, 1)(6, 3)(6, 5)

Now,

(i) A and B = AՌB = Փ

Since, There is no common events in Both A and B so the intersection is Null (Փ).

B or C = BUC

BUC = {(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5)(5, 6)(2, 1)(2, 2)(2, 3)(4, 1)}

B and C = BՌC

BՌC = {(1, 1), (1, 2)(1, 3)(1, 4)(3, 1)(3, 2)}

A and E = AՌE

AՌE = {(4, 6)(6, 4)(6, 5)(5, 6)(6, 6)}

A or F = AUF

AUF = {(2, 1)(2, 2)(2, 3)(2, 4)(2, 5)(2, 6)(4, 1)(4, 2)(4, 3)(4, 4)(6, 1)(6, 2)(6, 3)(6, 4)(6, 5)(6, 6)(1, 2)(1, 4)(1, 6)(3, 2)(3, 4)(3, 6)(4, 5)(5, 2)(5, 4)(5, 6)}

(ii)

(a) True, because AՌB = Փ

(b) True, because AՌB = Փ and AUB = S

(c) False, because AՌC = Փ

(d) False, because CՌD = Փ but CUD≠S

(e) True, because CՌDՌE = Փ and CUDUE = S

(f) True, because A’ՌB’ = Փ

(g) False, because AՌBՌF = Փ and AUBUF = S

Given: There are 4 slips in the box and mixed thoroughly.

To Find: Describe the given events .

Explanation: Here, Four slips of paper 1, 2, 3, and 4 are put in a box.

If Two slips are drawn from it one after the other without replacement. Then

The sample space for the experiment is:

S = {(1, 2)(1, 3), (1, 4)(2, 1)(2, 3)(2, 4)(3, 1)(3, 2)(3, 4)(4, 1)(4, 2)(4, 3)}

(i) A = number on the first slip is larger than the one on the second slip,

So, The sample space for A is:

{(2, 1)(3, 1)(3, 2)(4, 1)(4, 2)(4, 3)}

(ii) B = number on the second slip is greater than 2

So, The sample space for B is

{(1, 3)(2, 3)(1, 4)(2, 4)(3, 4)(4, 3)}

(iii) C = sum of the numbers on the two slips in 6 or 7

The sample space for C is

{(2, 4)(3, 4)(4, 2)(4, 3)}

(iv) D = number on the second slip is two times the number on the first slip

The sample space if:

{(1, 2)(2, 4)}

Now,

We can see, AՌD = Փ

Therefore, A and D are mutually exclusive events

Hence, A and D are mutually exclusive events.

Given: There is a deck of 52 playing card.

To Find: What is the sample space for the experiment and what is the event that the chosen card is black faced red.

Explanation: If a card is picked up from a deck of 52 playing cards, then the sample space of this experiment S is

(i) S = 52 cards in a deck

(ii) Let E is the event, and a black face card is chosen,

Then The possible outcomes in E are jack, queen, and king of clubs and the jack, queen and king of spades.

E = {J4, Q4, K4, Q*, KAM)

Hence, E = {J4, Q4, K4, Q*, KAM)

for each event to be a valid assignment of probability, the probability of each event in sample space should be less than 1 and the sum of probability of all the events should be exactly equal to 1

(i) it is valid as each P(w_i) (for i=1 to 7) lies between 0 to 1 and sum of P(w₁) =1

(ii) it is valid as each P(w_i) (for i=1 to 7) lies between 0 to 1 and sum of P(w₁) =1

(iii) it is not valid as sum of P(w_i)=2.8 which is greater than 1

(iv) it is not valid as which is greater than 1

given: die is thrown

Therefore, the total number of outcomes is six

n(S)=6

formula:

(i) let E be the event of getting a prime number

E= {2,3,5}

n(E)=3

(ii) let E be the event of getting 2 or 4

E= {2,4}

n(E)=2

(iii) let E be the event of getting a multiple of 2 or 3

E= {2,3,4,6}

n(E)=4

given: a pair of dice have been thrown, so the number of elementary events in sample space is 6²=36

formula:

therefore n(S)=36

(i) let E be the event that the sum 8 appears

E= {(2,6) (3,5) (4,4) (5,3) (6,2)}

n(E)=5

(ii) let E be the event of getting a doublet

E= {(1,1) (2,2) (3,3) (4,4) (5,5) (6,6)}

n(E)=6

(iii) let E be the event of getting a doublet of prime numbers

E= {((2,2) (3,3) (5,5)}

n(E)=3

(iv) let E be the event of getting a doublet of odd numbers

E= {(1,1) (3,3) (5,5)}

n(E)=3

(v) let E be the event of getting sum greater than 9

E= {(4,6) (5,5) (5,6) (6,4) (6,5) (6,6)}

n(E)=6

(vi) let E be the event of getting even on first die

E= {(2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}

n(E)=18

(vii) let E be the event of getting even on one and multiple of three on other

E= {(2,3) (2,6) (4,3) (4,6) (6,3) (6,6) (3,2) (3,4) (3,6) (6,2) (6,4)}

n(E)=11

(viii) let E be the event of getting neither 9 or 11 as the sum

E= {(3,6) (4,5) (5,4) (5,6) (6,3) (6,5)}

n(E)=6

(ix) let E be the event of getting sum less than 6

E= {(1,1) (1,2) (1,3) (1,4) (2,1) (2,2) (2,3) (3,1) (3,2) (4,1)}

n(E)=10

(x) let E be the event of getting sum less than 7

E= {(1,1) (1,2) (1,3) (1,4) (1,5) (2,1) (2,2) (2,3) (2,4) (3,1) (3,2) (3,3) (4,1) (4,2) (5,1)}

n(E)=15

(xi) let E be the event of getting more than 7

E= {(2,6) (3,5) (3,6) (4,4) (4,5) (4,6) (5,3) (5,4) (5,5) (5,6) (6,2) (6,3) (6,4) (6,5) (6,6)}

n(E)=15

(xii) let E be the event of getting neither a doublet nor a total of 10

E’ be the event that either a doublet or a sum of ten appears

E’= {(1,1) (2,2) (3,3) (4,6) (5,5) (6,4) (6,6) (4,4)}

n(E’) =8

Therefore P(E)=1-P(E’)

(xiii) let E be the event of getting odd number on first and 6 on second

E= {(1,6) (5,6) (3,6)}

n(E)=3

(xiv) let E be the event of getting greater than 4 on each die

E= {(5,5) (5,6) (6,5) (6,6)}

n(E)=4

(xv) let E be the event of getting total of 9 or 11

E= {(3,6) (4,5) (5,4) (5,6) (6,3) (6,5)}

n(E)=6

(xvi) let E be the event of getting total greater than 8

E= {(3,6) (4,5) (4,6) (5,4) (5,5) (5,6) (6,3) (6,4) (6,5) (6,6)}

n(E)=10

given: three dices are thrown

formula:

Total number of possible outcomes are 6³=216

Therefore n(S)=216

Let E be the event of getting total of 17 or 18

E= {(6,6,5) (6,5,6) (5,6,6) (6,6,6)}

n(E)=4

given: three coins are tossed together

formula:

Total number of possible outcomes are 2³=8

(i) let E be the event of getting exactly two heads

E= {(H, H, T) (H, T, H) (T, H, H)}

n(E)=3

(ii) let E be the event of getting at least two heads

E= {(H, H, T) (H, T, H) (T, H, H) (H, H, H)}

n(E)=4

(iii) let E be the event of getting at least one head and one tail

E= {(H, T, T) (T, H, T) (T, T, H) (H, H, T) (H, T, H) (T, H, H)}

n(E)=6

given: an ordinary year which includes 52 weeks and one day

formula:

so, we have to determine the probability of that one day being Sunday

Total number of possible outcomes are 7

Therefore n(S)=7

E= {M, T, W, T, F, S, SU}

n(E)=1

given: a leap year which includes 52 weeks and two days

formula:

so, we have to determine the probability of that remaining two days are Sunday and Monday

S= {MT, TW, WT, TF, FS, SSu, SuM}

Therefore n(S)=7

E= {SuM}

n(E)=1

given: bag which contains 8 red and 5 white balls

formula:

total number of ways of drawing three balls at random is ¹³C₃

therefore n(S)=286

(i) let E be the event of getting all white balls

E= {(W) (W) (W)}

n(E)= ⁵C₃=10

(ii) let E be the event of getting all red balls

E= {(R) (R) (R)}

n(E)= ⁸C₃=56

(iii) let E be the event of getting one red and two white balls

n(E)= ⁸C₁⁵C₂=80

given: three dice are rolled

formula:

so, we have to determine the probability of getting the same number on all the three dice

total number of possible outcomes are 6³=216

therefore n(S)=216

let E be the event of getting same number on all the three dice

E= {(1,1,1) (2,2,2) (3,3,3) (4,4,4) (5,5,5) (6,6,6)}

n(E)=6

given: three dice are rolled

formula:

so, we have to determine the probability of getting the sum of digits on dice greater than 10

total number of possible outcomes are 6²=36

therefore n(S)=36

let E be the event of getting same number on all the three dice

E= {(5,6) (6,5) (6,6)}

n(E)=3

given: pack of 52 cards

formula:

since a card is drawn from a pack of 52 cards, therefore number of elementary events in the sample space is

n(S)= ⁵²C₁ = 52

(i) let E be the event of drawing a black king

n(E)=²C₁ =2 (there are two black kings one of spade and other of club)

(ii) let E be the event of drawing a black card or a king

n(E)=²⁶C₁+⁴C₁-²C₁=28

we are subtracting 2 from total because there are two black king which are already counted and to avoid the error of considering it twice

(iii) let E be the event of drawing a black card and a king

n(E)=²C₁ =2 (there are two black kings one of spade and other of club)

(iv) let E be the event of drawing a jack, queen or king

n(E)=⁴C₁+⁴C₁+⁴C₁=12

(v) let E be the event of drawing neither a heart nor a king

now consider E’ as the event that either a heart or king appears

n(E’) =⁶C₁+⁴C₁-1=16 (there is a heart king so it is deducted)

P(E)=1-P(E’)

(vi) let E be the event of drawing a spade or king

n(E)=¹³C₁+⁴C₁-1=16

(vii) let E be the event of drawing neither an ace nor a king

now consider E’ as the event that either an ace or king appears

n(E’) =⁴C₁+⁴C₁=8

P(E)=1-P(E’)

(viii) let E be the event of drawing a diamond card

n(E)=¹³C₁=13

(ix) let E be the event of drawing not a diamond card

now consider E’ as the event that diamond card appears

n(E’) =¹³C₁=13

P(E)=1-P(E’)

(x) let E be the event of drawing a black card

n(E)=²⁶C₁=26 (spades and clubs)

(xi) let E be the event of drawing not an ace

now consider E’ as the event that ace card appears

n(E’) =⁴C₁=4

P(E)=1-P(E’)

(xii) let E be the event of not drawing a black card

n(E)=²⁶C₁=26 (red cards of hearts and diamonds)

given: pack of 52 cards from which 4 are dropped

formula:

we have to find the probability that the missing cards should be one from each suit

since from well shuffled pack of cards, 4 cards missed out total possible outcomes are

n(S)= ⁵²C₄=270725

let E be the event that four missing cards are from each suite

n(E)= ¹³C₁×¹³C₁×¹³C₁×¹³C₁=13⁴

given: deck of 52 cards

formula:

we have to find the probability that all the face cards of same suits are drawn

total possible outcomes are

n(S)=⁵²C₄

let E be the event that all the cards drawn are face cards of same suit

n(E)=4×⁴C₄=4

given: numbered tickets from 1 to 20

formula:

to find the probability of the ticket drawn having a number which is a multiple of 3 or 7

Since one ticket is drawn from a lot of mixed number, total possible outcomes are

n(S)=²⁰C₁=20

let E be the event of getting ticket which has number that is multiple of 3 or 7

E= {3,6,9,12,15,18,7,14}

n(E)=8

given: 6 red, 4 white and 8 blue balls

formula:

three balls are drawn so, we have to find the probability that one is red, one is white and one is blue

total number of outcomes for drawing 3 balls are ¹⁸C₃

Therefore n(S)= ¹⁸C₃=816

Let E be the event that one red, one white and one blue ball is drawn

n(E)= ⁶C₁⁴C₁⁸C₁=192

given: bag which contains 4 red, 5 black and 7 white balls

formula:

two balls are drawn at random, therefore

total possible outcomes are ¹⁶C₂

therefore n(S)=120

(i) let E be the event of getting both white balls

E= {(W) (W)}

n(E)= ⁷C₂=21

(ii) let E be the event of getting one black and one red ball

E= {(B) (R)}

n(E)= ⁵C₁⁴C₁=20

(iii) let E be the event of getting both balls of same colour

E= {(B) (B)} or {(W) (W)} or {(R) (R)}

n(E)= ⁷C₂+⁵C₂+⁴C₂=37

given: bag which contains 6 red, 8 blue and 4 white balls

formula:

two balls are drawn at random, therefore

total possible outcomes are ¹⁸C₃

therefore n(S)=816

(i) let E be the event of getting one red and two white balls

E= {(W) (W) (R)}

n(E)= ⁶C₁⁴C₂=36

(ii) let E be the event of getting two blue and one red

E= {(B) (B) (R)}

n(E)= ⁸C₂⁶C₁=168

(iii) let E be the event that one of the balls must be red

E= {(R) (B) (B)} or {(R) (W) (W)} or {(R) (B) (W)}

n(E)= ⁶C₁⁴C₁⁸C₁+⁶C₁⁴C₂+⁶C₁⁸C₂=396

given: pack of 52 playing cards

formula:

five cards are drawn at random, therefore

total possible outcomes are ⁵²C₅

therefore n(S)=2598960

(i) let E be the event that exactly only one ace is present

n(E)= ⁴C₁⁴⁸C₄=778320

(ii) let E be the event that at least one ace is present

E= {1 or 2 or 3 or 4 ace(s)}

n(E)= ⁴C₁⁴⁸C₄+⁴C₂⁴⁸C₃+⁴C₃⁴⁸C₂+⁴C₄⁴⁸C₁=886656

given: pack of 52 cards from which face cards are removed

formula:

four cards are drawn from the remaining 40 cards, so we have to find the probability that all of them belong to different suit

total possible outcomes of drawing four cards are ⁴⁰C₄

therefore n(S)= ⁴⁰C₄

let E be the event that 4 cards belong to different suit

n(E)= ¹⁰C₁¹⁰C₁¹⁰C₁¹⁰C₁=10000

given: four men and six women

formula:

from the city council one person is selected as a council member so, we have to find the probability that it is a woman

total possible outcomes of selecting a person is ¹⁰C₁

therefore n(S)= ¹⁰C₁

let E be the event that it is a woman

n(E)= ⁶C₁=6

given: box with 100 bulbs of which, 20 are defective

formula:

ten bulbs are drawn at random for inspection, therefore

total possible outcomes are ¹⁰⁰C₁₀

therefore n(S)= ¹⁰⁰C₁₀

(i) let E be the event that all ten bulbs are defective

n(E)= ²⁰C₁₀

(ii) let E be the event that all ten good bulbs are selected

n(E)= ⁸⁰C₁₀

(iii) let E be the event that at least one bulb is defective

E= {1,2,3,4,5,6,7,8,9,10} where 1,2,3,4,5,6,7,8,9,10 are the number of defective bulbs

Let E’ be the event that none of the bulb is defective

n(E’) = ⁸⁰C₁₀

Therefore,

P(E)=1-P(E’)

(iv) let E be the event that none of the selected bulb is defective

n(E)= ⁸⁰C₁₀

given: word “SOCIAL”

formula:

In the random arrangement of the alphabets of word “SOCIAL” we have to find the probability that vowels come together

total possible outcomes of arranging the alphabets are 6!

therefore n(S)=6!

let E be the event that vowels come together

number of vowels in SOCIAL is A, I, O

therefore, number of ways to arrange them so (A, I, O) come together

n(E)= 4! × 3!

given: word “CLIFTON”

formula:

In the random arrangement of the alphabets of word “CLIFTON” we have to find the probability that vowels come together

total possible outcomes of arranging the alphabets are 7!

therefore n(S)=7!

let E be the event that vowels come together

number of vowels in SOCIAL is I, O

therefore, number of ways to arrange them so (I, O) come together

n(E)= 6! × 2!

given: word “FORTUNATES”

formula:

In the random arrangement of the alphabets of word “FORTUNATES” we have to find the probability that two T’s come together

total possible outcomes of arranging the alphabets are 10!

therefore n(S)=10!

let E be the event that T’s come together

therefore, number of ways to arrange them so (T, T) come together

n(E)= 9! × 2!

given: two men and two women

formula:

committee of two persons is to be formed from two men and two women, therefore

total possible outcomes of selecting two persons is ⁴C₂

therefore n(S)=6

(i) let E be the event that no man is in the committee

E= {W, W}

n(E)= ²C₂=1 (only woman)

(ii) let E be the event that one man is present in committee

E= {M W}

n(E)= ²C₁²C₁=4

(iii) let E be the event that two men is in the committee

E= {M, M}

n(E)= ²C₂=1 (only men)

given: odds in favour of event is 2:3

formula:

we have to find the probability of occurrence of this event

total possible outcomes is 2k+3k=5k

therefore n(S)= 5k

let E be the event that it occurs

n(E)=2k

probability of occurrence is

given: odds against of event is 7:9

formula:

we have to find the probability of non-occurrence of this event

total possible outcomes are 7k+9k=16k

therefore n(S)= 16k

let E be the event that it occurs

n(E)=9k

probability of occurrence is

Therefore, the probability of non-occurrence of the event is

given: 2 white, 3 red, 5 green, 4 black

formula:

two balls are drawn one by one, we have to find the probability that they are of different colours

total possible outcomes are ¹⁴C₂

therefore n(S)= ¹⁴C₂=91

let E be the event that all balls are of same colour

E= {WW, RR, GG, BB}

n(E)= ²C₂+³C₂+⁵C₂+⁴C₂=20

probability of occurrence is

Therefore, the probability of non-occurrence of the event (all balls are different) is

given: two unbiased dice are thrown

formula:

total possible outcomes of from the dice is ⁶C₁⁶C₁

therefore n(S)=36

(i) let E be the event that neither a doublet nor a total of 8 will appear

E’ be the event that a doublet or a total of 8 occurs

E’= {(1,1) (2,2) (3,3) (4,4) (5,5) (6,6) (2,6) (6,2) (3,5) (5,3)}

n(E’) = 10

Therefore P(E) is

(ii) let E be the event that sum of number obtain on the dice is neither a multiple of 2 or 3

E’ be the event that sum of number obtain on the dice is either a multiple of 2 or 3, that is total should be 2,3,4,6,8,9,10,12

E’= {(1,1) (1,2) (2,1) (1,3) (2,2) (3,1) (1,5) (2,4) (3,3) (4,2) (5,1) (2,6) (3,5) (4,4) (5,3) (6,2) (3,6) (4,5) (5,4) (6,3) (4,6) (5,5) (6,4) (6,6)}

n(E’) = 24

Therefore P(E) is

given: bag which contains 8 red, 3 white, 9 blue balls

formula:

three balls are drawn at random therefore

total possible outcomes of selecting two persons is ²⁰C₃

therefore n(S)=²⁰C₃=1140

(i) let E be the event that all the balls are blue

E= {B, B, B}

n(E)= ⁹C₃=84

(ii) let E be the event that all balls are of different colour

E= {B W R}

n(E)= ⁸C₁³C₁⁹C₁=216

given: bag which contains 5 red, 6 white, 7 black balls

formula:

two balls are drawn at random

total possible outcomes are ¹⁸C₂

therefore n(S)= ¹⁸C₂=153

let E be the event that both balls are either red or black

n(E)=⁵C₂+⁷C₂=31

probability of occurrence of the event is

given: a letter is chosen at random from English alphabet

formula:

total possible outcomes of selecting an alphabet is ²⁶C₁

therefore n(S)=²⁶C₁=26

(i) let E be the event that a vowel has been chosen

E= {a, e, i, o, u}

n(E)= ⁵C₁=5

(ii) let E be the event that a consonant is chosen

E= all alphabets - {a, e, i, o, u}

n(E)= ²¹C₁=21

given: six numbers are chosen from 1-20

formula:

we have to find the probability of winning the prize

total possible outcomes of selecting six numbers from 1-20 is ²⁰C₆

therefore n(S)= ²⁰C₆=38760

let E be the event that all six numbers match with the given number (as winning number is fixed)

n(E)=1

probability of occurrence of the event is

given: 20 cards numbered from 1-20

formula:

one card is drawn at random therefore total possible outcomes are ²⁰C₁

therefore n(S)=²⁰C₁=20

(i) let E be the event that the number on the drawn card is a multiple of 4

E= {4, 8, 12, 16, 20}

n(E)= ⁵C₁=5

(ii) let E be the event that the number on the drawn card is not a multiple of 4

E’ be the event that the number on the drawn card is a multiple of 4

E’= {4, 8, 12, 16, 20}

n(E)= ⁵C₁=5

P(E)=1-P(E’)

(iii) let E be the event that the number on the drawn card is odd

E= {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}

n(E)= ¹⁰C₁=10

(iv) let E be the event that the number on the drawn card is greater than 12

E= {13, 14, 15, 16, 17, 18, 19, 20}

n(E)= ⁸C₁=8

(v) let E be the event that the number on the drawn card is a multiple of 5

E= {5, 10, 15, 20}

n(E)= ⁴C₁=4

(vi) let E be the event that the number on the drawn card is not divisible by 6

let E’ be the event that number on the drawn card is divisible by 6

E’= {6, 12, 18}

n(E’)= ³C₁=3

P(E)=1-P(E’)

given: two dices are thrown

formula:

total possible outcomes are ⁶C₁⁶C₁

therefore n(S)=6²=36

(i) let E be the event that total sum is 4 on dice

E= {(1,3) (3,1) (2,2)}

n(E)= ³C₁=3

Therefore, probability of event E’ is

P(E’) =1-P(E)

Odds in favour of getting sum as 4 is P(E):P(E’) =1:11

(ii) let E be the event that total sum is 5 on dice

E= {(1,4) (2,3) (3,2) (4,1)}

n(E)= ⁴C₁=4

Therefore, probability of event E’ is

P(E’) =1-P(E)

Odds in favour of getting sum as 5 is P(E):P(E’) =1:8

(iii) let E be the event that total sum is 6 on dice

E= {(1,5) (2,4) (3,3) (4,2) (5,1)}

n(E)= ⁵C₁=5

Therefore, probability of event E’ is

P(E’) =1-P(E)

Odds against of getting sum as 6 is P(E’):P(E) =31:5

given: pack of 52 playing cards

formula:

one card is drawn from the given pack of cards, total possible outcomes are ⁵²C₁

therefore n(S)=⁵²C₁=52

(i) let E be the event of getting a spade

n(E)= ¹³C₁=13

Therefore, probability of event E’ is

P(E’) =1-P(E)

Odds in favour of getting a spade is P(E):P(E’) =1:3

(ii) let E be the event of getting a king

n(E)= ⁴C₁=4

Therefore, probability of event E’ is

P(E’) =1-P(E)

Odds in favour of getting a king is P(E):P(E’) =1:12

given: box containing 10 red, 20 blue and 30 green marbles

formula:

five marbles are drawn from the given box, total possible outcomes are ⁶⁰C₅

therefore n(S)=⁶⁰C₅

(i) let E be the event of getting all blue balls

n(E)= ²⁰C₅

(ii) let E be the event of getting at least one green

let E’ be the event of getting no green

E’= {(5B) (1R 4B) (2R 3B) (3R 2B) (4R 1B) (5R)}

5B=²⁰C₅

1R 4B=¹⁰C₁²⁰C₄

2R 3B=¹⁰C₂²⁰C₃

3R 2B=¹⁰C₃²⁰C₂

4R 1B=¹⁰C₄²⁰C₁

5R= ¹⁰C₅

P(E) =1-P(E’)

given: box containing 6 red marbles numbered 1-6, 4 white marbles numbered 12-15

formula:

one marble is drawn from the given box, total possible outcomes are ¹⁰C₁

therefore n(S)=¹⁰C₁=10

(i) let E be the event of getting white marble

n(E)= ⁴C₁=4

(ii) let E be the event of getting white marble with odd numbered

E= {13,15}

n(E)= 2

(iii) let E be the event of getting even numbered marble

E= {2, 4, 6, 12, 24}

n(E)= 5

(iv) let E₁ be the event of getting red marble

(from (i))

Let E₂ be the event of getting even numbered marble

(from (ii))

Therefore (E₁ꓵ E₂) = red coloured and even numbered

n (E₁ꓵ E₂) =3

By law of addition P(E₁∪ E₂) = P(E₁)+P(E₂)- P(E₁ꓵ E₂)

given: class consisting of 10 boys and 8 girls

formula:

three students are selected at random, total possible outcomes are ¹⁸C₃

therefore n(S)=¹⁸C₃=816

(i) let E be the event that all are boys

n(E)= ¹⁰C₃=120

(ii) let E be the event that all are girls

n(E)= ⁸C₃=56

(iii) let E be the event that one boy and two girls are selected

n(E)= ⁸C₁¹⁰C₂=360

(iv) let E be the event that at least one girl is in the group

E= {1,2,3}

n(E)= ⁸C₁¹⁰C₂+⁸C₂¹⁰C₁+⁸C₃¹⁰C₀=696

(v) let E be the event that at most one girl is in the group

E= {0, 1}

n(E)= ⁸C₀¹⁰C₃+⁸C₁¹⁰C₂=480

given: pack of 52 playing cards

formula:

five cards are drawn from the given deck

we have to find the probability that all of them are hearts

total possible outcomes of one card from the pack is ⁵²C₁

therefore n(S)= ⁵²C₅=2598960

let E be the event that all cards belong to hearts

n(E)=¹³C₅=1287

probability of occurrence of the event is

given: bag containing tickets numbered 1-20

formula:

two tickets are drawn at random, total possible outcomes are ²⁰C₂

therefore n(S)=²⁰C₂=190

(i) let E be the event that both tickets have prime number

E= {2,3,5,7,11,13,17,19}

n(E)= ⁸C₂=28

(ii) let E be the event that one ticket has prime number and other is a multiple of 4

E= {2,3,5,7,11,13,17,19} for prime number

E= {4,8,12,16,20} for multiple of 4

n(E)= ⁸C₁⁵C₁=40

given: urn containing 7 white, 5 black and 3 red

formula:

two balls are drawn at random, total possible outcomes are ¹⁵C₂

therefore n(S)=¹⁸C₂=105

(i) let E be the event that both balls are red

n(E)= ³C₂=3

(ii) let E be the event that one is red and other is black

n(E)= ³C₁⁵C₁=15

(iii) let E be the event that one ball is white

n(E)= ⁸C₁⁷C₁=56

given: pair of dice

formula:

A and B throw a pair of dice

we have to find the probability that B throw’s a higher number

total possible outcomes are ⁶C₁⁶C₁

therefore n(S)= 6²=36

let E be the event that A throws 9, and B throws greater than 9

E= {(4,6) (5,5) (5,6) (6,4) (6,5) (6,6)}

n(E)= 6

probability of occurrence of the event is

given: game of whist is being played

formula:

we have to find the probability that all four kings are held by a specific player

since in one hand at whist a player has 13 cards

total possible outcomes are ⁵²C₁₃

therefore n(S)= ⁵²C₁₃

let E be the event that a player has 4 kings

n(E)= ⁴⁸C₉⁴C₄

probability of occurrence of this event is

given: word “UNIVERSITY”

formula:

we have to find the probability that two I’s do not come together

total possible outcomes for arrangement of alphabets are 10!

therefore n(S)= 10!

let E be the event that both I’s come together

n(E)= 2! × 9!

probability of occurrence of this event is

Let E’ be the event that both I’s do not come together

Therefore, the probability that two I’s do not come together is

P(E’) =1-P(E)

Given A and B are two mutually exclusive events

And, P(A) = 0.4 P(B) = 0.5

By definition of mutually exclusive events we know that:

P(A ∪ B) = P(A) + P(B)

We have to find-

i) P(A ∪ B) = P(A) + P(B) = 0.5 + 0.4 = 0.9

ii) P(A’ ∩ B’) = P(A ∪ B)’ {using De Morgan’s Law}

⇒ P(A’ ∩ B’) = 1 – P(A ∪ B) = 1 – 0.9 = 0.1

iii) P(A’ ∩ B) = This indicates only the part which is common with B and not A ⇒ This indicates only B.

P(only B) = P(B) – P(A ∩ B)

As A and B are mutually exclusive So they don’t have any common parts ⇒ P(A ∩ B) = 0

∴ P(A’ ∩ B) = P(B) = 0.5

iv) P(A ∩ B’) = This indicates only the part which is common with A and not B ⇒ This indicates only A.

P(only A) = P(A) – P(A ∩ B)

As A and B are mutually exclusive So they don’t have any common parts ⇒ P(A ∩ B) = 0

∴ P(A ∩ B’) = P(A) = 0.4

Given A and B are two events

And, P(A) = 0.54 P(B) = 0.69 P(A ∩ B) = 0.35

By definition of P(A or B) under axiomatic approach we know that:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

We have to find-

i) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 0.54 + 0.69 – 0.35 = 0.88

ii) P(A’ ∩ B’) = P(A ∪ B)’ {using De Morgan’s Law}

⇒ P(A’ ∩ B’) = 1 – P(A ∪ B) = 1 – 0.88 = 0.12

iii) P(A ∩ B’) = This indicates only the part which is common with A and not B ⇒ This indicates only A.

P(only A) = P(A) – P(A ∩ B)

∴ P(A ∩ B’) = P(A) - P(A ∩ B) = 0.54 – 0.35 = 0.19

iv) P(A’ ∩ B) = This indicates only the part which is common with B and not A ⇒ This indicates only B.

P(only B) = P(B) – P(A ∩ B)

∴ P(A’ ∩ B) = P(B) – P(A ∩ B) = 0.69 – 0.35 = 0.34

We need to fill the table:

i) By definition of P(A or B) under axiomatic approach we know that:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

Using data from table, we get:

∴ P(A ∪ B) =

ii) By definition of P(A or B) under axiomatic approach we know that:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

⇒ P(B) = P(A ∪ B) + P(A ∩ B) – P(A)

Using data from table, we get:

∴ P(B) = 0.6 + 0.25 – 0.35 = 0.5

iii) By definition of P(A or B) under axiomatic approach we know that:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

⇒ P(A ∩ B) = P(B) + P(A) - P(A ∪ B)

Using data from table, we get:

∴ P(A ∩ B) = 0.5 + 0.35 – 0.7 = 0.15

Filled table is:

Given A and B are two events

And, P(A) = 0.3 P(B) = 0.5 P(A ∪ B) = 0.5

We need to find P(A ∩ B).

By definition of P(A or B) under axiomatic approach(also called addition theorem) we know that:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

∴ P(A ∩ B) = P(A) + P(B) – P(A ∪ B)

⇒ P(A ∩ B) = 0.3 + 0.4 – 0.5 = 0.7 – 0.5 = 0.2

∴ P(A ∩ B) = 0.2

Given A and B are two events

And, P(A) = 0.5 P(B) = 0.3 P(A ∩ B) = 0.2

We need to find P(A ∪ B).

By definition of P(A or B) under axiomatic approach(also called addition theorem) we know that:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

⇒ P(A ∪ B) = 0.5 + 0.3 – 0.2 = 0.8 – 0.2 = 0.6

∴ P(A ∪ B) = 0.6

Given A and B are two events

And, P(A’) = 0.5 P(A ∩ B) = 0.3 P(A ∪ B) = 0.8

∵ P(A’) = 1 – P(A) ⇒ P(A) = 1 – 0.5 = 0.5

We need to find P(B).

By definition of P(A or B) under axiomatic approach(also called addition theorem) we know that:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

∴ P(B) = P(A ∪ B) + P(A ∩ B) – P(A)

⇒ P(B) = 0.8 + 0.3 – 0.5 = 1.1 – 0.5 = 0.6

∴ P(B) = 0.6

Given A and B are two mutually exclusive events

And, P(A) = 1/2 P(B) = 1/3

We need to find P(A ‘or’ B).

P(A or B) = P(A ∪ B)

By definition of mutually exclusive events we know that:

P(A ∪ B) = P(A) + P(B)

∴ P(A ∪ B) = P(A) + P(B)

= 1/2 + 1/3 = 5/6

As, out of 3 events A,B and C only one can happen at a time which means no event have anything common.

∴ We can say that A , B and C are mutually exclusive events.

By definition of mutually exclusive events we know that:

P(A ∪ B ∪ C) = P(A) + P(B) + P(C)

According to question one event must happen.

This implies A or B or C is a sure event.

∴ P(A ∪ B ∪ C) = 1 …Equation 1

We need to find odd against C

Given,

Odd against A = 8/3

⇒

⇒

⇒ 8 P(A) = 3 – 3 P(A)

⇒ 11 P(A) = 3

∴ P(A) = …Equation 2

Similarly, we are given with: Odd against B = 5/2

⇒

⇒

⇒ 5 P(B) = 2 – 2 P(B)

⇒ 7 P(B) = 2

∴ P(B) = …Equation 3

From equation 1,2 and 3 we get:

P(C) = 1 - =

∴ P(C’) = 1 – (34/77) = 43/77

∴ Odd against C =

Let A and B are two events.

As, out of 2 events A and B only one can happen at a time which means no event have anything common.

∴ We can say that A and B are mutually exclusive events.

By definition of mutually exclusive events we know that:

P(A ∪ B) = P(A) + P(B)

According to question one event must happen.

This implies A or B is a sure event.

∴ P(A ∪ B) = P(A) + P(B) = 1 …Equation 1

Given, P(A) = (2/3)P(B)

To find: odds in favour of B

∴

⇒

∴ P(B’) = 1 – 3/5 = 2/5

∴ Odd in favour of B =

As a card is drawn from a deck of 52 cards

Let S denotes the event of card being a spade and K denote the event of card being King.

As we know that a deck of 52 cards contains 4 suits (Heart ,Diamond ,Spade and Club) each having 13 cards. The deck has 4 king cards one from each suit.

We know that probability of an event E is given as-

P(E) =

Where n(E) = numbers of elements in event set E

And n(S) = numbers of elements in sample space.

Hence,

P(S) =

P(K) =

And P(S ∩ K) =

We need to find the probability of card being spade or king, i.e.

P(Spade ‘or’ King) = P(S ∪ K)

Note: By definition of P(A or B) under axiomatic approach(also called addition theorem) we know that:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

∴ P(S ∪ K) = P(S) + P(K) - P(S ∩ K)

⇒ P(S ∪ K) =

∴ P(S ∪ K) = 4/13

In a single throw of 2 die, we have total 36(6 × 6) outcomes possible.

Say, n(S) = 36 where S represents Sample space

Let A denotes the event of getting a doublet.

∴ A = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

∴ P(A) =

And B denotes the event of getting a total of 9

∴ B = {(3,6), (6,3), (4,5), (5,4)}

P(B) =

We need to find probability of the event of getting neither a doublet nor a total of 9.

P(A’ ∩ B’) = ?

As, P(A’ ∩ B’) = P(A ∪ B)’ {using De Morgan’s theorem}

P(A’ ∩ B’) = 1 – P(A ∪ B)

Note: By definition of P(E or F) under axiomatic approach(also called addition theorem) we know that:

P(E ∪ F) = P(E) + P(F) – P(E ∩ F)

∴ P(A ∪ B) = {As P(A ∩ B) = 0 since nothing is common in set A and B ⇒ n(A ∩ B) = 0 }

Hence,

P(A’ ∩ B’) = 1 – (5/18) = 13/18

Given, Sample space is the set of first 500 natural numbers.

∴ n(S) = 500

Let A be the event of choosing the number such that it is divisible by 3

∴ n(A) = [500/3] = [166.67] = 166 {where [.] represents Greatest integer function}

∴ P(A) =

Let B be the event of choosing the number such that it is divisible by 5

∴ n(B) = [500/5] = [100] = 100 {where [.] represents Greatest integer function}

∴ P(B) =

We need to find the P(such that number chosen is divisible by 3 or 5)

∵ P(A or B) = P(A ∪ B)

Note: By definition of P(E or F) under axiomatic approach(also called addition theorem) we know that:

P(E ∪ F) = P(E) + P(F) – P(E ∩ F)

∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

We don’t have value of P(A ∩ B) which represents event of choosing a number such that it is divisible by both 3 and 5 or we can say that it is divisible by 15.

n(A ∩ B) = [500/15] = [33.34] = 33

∴ P(A ∩ B) =

∴ P(A ∪ B) =

If a dice is thrown twice , it has a total of(6 × 6) 36 possible outcomes.

If S represents the sample space then,

n(S) = 36

Let A represent events the event such that 3 comes in the first throw.

∴ A = {(1,3), (2,3), (3,3), (4,3), (5,3), (6,3)}

⇒ P(A) =

Let B represent events the event such that 3 comes in the second throw.

∴ B = {(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)}

⇒ P(B) =

Clearly (3,3) is common in both events-

∴ P(A ∩ B) =

We need to find the probability of event such that at least one of the 2 throws give 3 i.e. P(A or B) = P(A ∪ B)

Note: By definition of P(E or F) under axiomatic approach(also called addition theorem) we know that:

P(E ∪ F) = P(E) + P(F) – P(E ∩ F)

∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

⇒ P(A ∪ B) =

Hence,

P(at least one of the two throws comes to be 3) =

As a card is drawn from a deck of 52 cards

Let S denotes the event of card being a spade and K denote the event of card being Ace.

As we know that a deck of 52 cards contains 4 suits (Heart ,Diamond ,Spade and Club) each having 13 cards. The deck has 4 Ace cards one from each suit.

We know that probability of an event E is given as-

P(E) =

Where n(E) = numbers of elements in event set E

And n(S) = numbers of elements in sample space.

Hence,

P(S) =

Similarly,

P(K) =

And P(S ∩ K) =

We need to find the probability of card being spade or king, i.e.

P(Spade ‘or’ Ace) = P(S ∪ K)

Note: By definition of P(A or B) under axiomatic approach(also called addition theorem) we know that:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

∴ P(S ∪ K) = P(S) + P(K) - P(S ∩ K)

⇒ P(S ∪ K) =

∴ P(S ∪ K) = 4/13

Let E denotes the event that student passes in English examination.

And H be the event that student passes in Hindi exam.

Given,

P(E) = 0.75

P(passing both) = P(E ∩ H) = 0.5

P(passing neither) = P(E’ ∩ H’) = 0.1

P(H) = ?

As, we know that P(A’ ∩ B’) = P(A ∪ B)’ {using De Morgan’s law}

∴ P(E’ ∩ H’) = P(E ∪ H)’

⇒ 0.1 = 1 – P(E ∪ H)

⇒ P(E ∪ H) = 1 – 0.1 = 0.9

Note: By definition of P(A or B) under axiomatic approach(also called addition theorem) we know that:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

∴ P(E ∪ H) = P(E) + P(H) – P(E ∩ H)

⇒ 0.9 = 0.75 + P(H) – 0.5

⇒ 1.4 – 0.75 = P(H)

∴ P(H) = 0.65

Given, Sample space is the set of first 100 natural numbers.

∴ n(S) = 100

Let A be the event of choosing the number such that it is divisible by 4

∴ n(A) = [100/4] = [25] = 25 {where [.] represents Greatest integer function}

∴ P(A) =

Let B be the event of choosing the number such that it is divisible by 6

∴ n(B) = [100/6] = [16.67] = 16 {where [.] represents Greatest integer function}

∴ P(B) =

We need to find the P(such that number chosen is divisible by 4 or 6)

∵ P(A or B) = P(A ∪ B)

Note: By definition of P(E or F) under axiomatic approach(also called addition theorem) we know that:

P(E ∪ F) = P(E) + P(F) – P(E ∩ F)

∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

We don’t have value of P(A ∩ B) which represents event of choosing a number such that it is divisible by both 4 and 6 or we can say that it is divisible by 12.

n(A ∩ B) = [100/12] = [8.33] = 8

∴ P(A ∩ B) =

∴ P(A ∪ B) =

In a deck of 52 cards there are 2 colours. Each colour having 26 cards.

As we need to choose 4 cards out of 52. Let S represents the sample space.

∴ n(S) = ⁵²C₄

Let A represents the event that all 4 cards drawn are black.

∴ n(A) = ways in which 4 cards can be selected from 26 black cards.

⇒ n(A) = ²⁶C₄

∴ P(A) = ²⁶C₄/⁵²C₄ =

Let B represents the event that all 4 cards drawn are red.

∴ n(B) = ways in which 4 cards can be selected from 26 red cards.

⇒ n(B) = ²⁶C₄

∴ P(B) = ²⁶C₄/⁵²C₄ =

As we need to find the probability of event such that all drawn cards are from same colour. This means we need to find

P(A ∪ B)

Note: By definition of P(E or F) under axiomatic approach(also called addition theorem) we know that:

P(E ∪ F) = P(E) + P(F) – P(E ∩ F)

∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

As both events A and B have no common elements or we can say that they are mutually exclusive

∴ P(A ∩ B) = 0

Hence,

P(A ∪ B) = P(A) + P(B) =

Let E denotes the event that student passed in first examination.

And H be the event that student passed in second exam.

S is the sample space containing the students who appeared for the exam.

Given,

n(S) = 100

n(E) = 60

n(H) = 50

also no of students who passed both exam = n(E ∩ H) = 30

∴ P(E) =

Similarly, P(H) =

And, P(E ∩ H) =

We need to find the probability of event such that a student selected at random has passed at least one examination.

This can be given as – P(E or H) = P(E ∪ H)

Note: By definition of P(A or B) under axiomatic approach(also called addition theorem) we know that:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

∴ P(E ∪ H) = P(E) + P(H) – P(E ∩ H)

⇒ P(E ∪ H) =

∴ P(E ∪ H) = 4/5

As box contains 26 balls. Let S represents the sample space.

∴ n(S) = 26

Let W denotes the event of drawing a white ball and R represents event of drawing red ball.

As, n(W) = 10 and n(R) = 6

∴ P(W) = 10/26 = 5/13

And P(R) = 6/26 = 3/13

As both events have nothing in common so we can say that they are mutually exclusive.

We need to find the probability of the event such that ball drawn is red or white.

P(Red or White) = P(R ∪ W)

As events are mutually exclusive

∴ P(R ∪ W) = P(R) + P(W)

⇒ P(R ∪ W) = 3/13 + 5/13 = 8/13

Given, odds in favour of A is

⇒

⇒ 1 – P(A) = 3P(A)

⇒ 4P(A) = 1 ⇒ P(A) = 1/4

Odds in favour of horse B is

⇒

⇒ 1 – P(B) = 4P(B)

⇒ 5P(B) = 1 ⇒ P(B) = 1/5

Odds in favour of horse C is

⇒

⇒ 1 – P(C) = 5P(C)

⇒ 6P(C) = 1 ⇒ P(C) = 1/6

Odds in favour of horse D is

⇒

⇒ 1 – P(D) = 6P(D)

⇒ 7P(D) = 1 ⇒ P(D) = 1/7

We have to find the probability that one of the horses win the race.

∵ only one horse can win the race ⇒ A ,B,C and D are mutually exclusive events.

We need to find P(A ∪ B ∪ C ∪ D).

∵ A ,B,C and D are mutually exclusive events.

∴ P(A ∪ B ∪ C ∪ D) = P(A) + P(B) + P(C) + P(D)

=

Hence,

probability that one of the horses win the race = 319/420

Let T denotes the event that person travels by train and A denotes that event that person travels by plane

Given, P(T) = 3/5 and P(A) = 1/4

We need to find the probability that person travels by plane or train i.e. P(T or A) = P(T ∪ A)

Note: By definition of P(A or B) under axiomatic approach(also called addition theorem) we know that:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

∴ P(T ∪ A) = P(T) + P(A) – P(A ∩ T)

∵ A person can never travel with both plane and train simultaneously.

∴ P(A ∩ T) = 0

Hence,

P(T ∪ A) = P(T) + P(A) = 3/5 + 1/4 = 17/20

As 2 cards are drawn from a deck of 52 cards. This can be done in ⁵²C₂ ways. If S represents the sample space,

n(S) = ⁵²C₂

Let B represents the event that both drawn cards are black.

∵ A deck of 52 cards has 26 black cards. So 2 cards can be selected out of those 26 in ²⁶C₂ ways

∴ n(B) = ²⁶C₂

∴ P(B) =

Let K represents the event that both drawn cards are king.

∵ A deck of 52 cards has 4 king cards. So 2 cards can be selected out of those 4 in ⁴C₂ ways

∴ n(K) = ⁴C₂

∴ P(K) =

We need to find the probability that either both are black or both are kings i.e. P(B or K) = P(B ∪ K)

Note: By definition of P(A or B) under axiomatic approach(also called addition theorem) we know that:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

∴ P(B ∪ K) = P(B) + P(K) – P(B ∩ K)

As we don’t have value of P(B ∩ K) so we will find it first.

As there is a common element among the events B and K as both the cards can be a king and can be black 2.

∵ 2 black king cards are present so we need to select 2 cards oout of them only. This can be done in ²C₂ ways = 1

∴ P(B ∩ K) =

∴ P(B ∪ K) =

Let E denotes the event that student passed in first examination.

And H be the event that student passed in second exam.

Given, P(E) = 0.8 and P(H) = 0.7

Also probability of passing atleast one exam i.e P(E or H) = 0.95

Or, P(E ∪ H) = 0.95

We have to find the probability of the event in which students pass both the examinations i.e. P(E ∩ H)

Note: By definition of P(A or B) under axiomatic approach(also called addition theorem) we know that:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

∴ P(E ∪ H) = P(E) + P(H) – P(E ∩ H)

⇒ P(E ∩ H) = P(E) + P(H) – P(E ∪ H)

⇒ P(E ∩ H) = 0.7 + 0.8 – 0.95 = 1.5 – 0.95 = 0.55

∴ Probability of passing both the exams = P(E ∩ H) = 0.55

Let S represents the sample space containing all possible ways of selecting 2 items (out 0f 70 nuts and bolts)

Let R represents the event of drawing 2 rusted item.

And B be the event of drawing 2 bolts

According to question-

n(S) = total ways of selecting 2 items out of 70 = ⁷⁰C₂

∵ half of bolts (30/2 = 15) and half of nuts(40/2 = 20) are rusted

∴ n(R) = no of ways in which 2 rusted items can be drawn.

⇒ n(R) = ³⁵C₂

And n(B) = no of ways in which 2 bolts can be drawn.

∴ n(B) = ³⁰C₂

Also, n(B ∩ R) = no of ways of selecting 2 items such that they are rusted bolts

∴ n(B ∩ R) = ¹⁵C₂

Hence, we have –

P(R) =

P(B) =

And P(R ∩ B) =

As we have to find probability of the events such that drawn items are rusted or bolts i.e. P(R ∪ B)

Note: By definition of P(A or B) under axiomatic approach(also called addition theorem) we know that:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

∴ P(R ∪ B) = P(R) + P(B) – P(R ∩ B)

⇒ P(R ∪ B) =

Given, Sample space is the set of first 200 natural numbers.

∴ n(S) = 200

Let A be the event of choosing the number such that it is divisible by 6

∴ n(A) = [200/6] = [33.334] = 33 {where [.] represents Greatest integer function}

∴ P(A) =

Let B be the event of choosing the number such that it is divisible by 8

∴ n(B) = [200/8] = [25] = 25 {where [.] represents Greatest integer function}

∴ P(B) =

We need to find the P(such that number chosen is divisible by 6 or 8)

∵ P(A or B) = P(A ∪ B)

Note: By definition of P(E or F) under axiomatic approach(also called addition theorem) we know that:

P(E ∪ F) = P(E) + P(F) – P(E ∩ F)

∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

We don’t have value of P(A ∩ B) which represents event of choosing a number such that it is divisible by both 4 and 6 or we can say that it is divisible by 24.

n(A ∩ B) = [200/24] = [8.33] = 8

∴ P(A ∩ B) =

∴ P(A ∪ B) =

When a coin is tossed 4 times. A total of 2⁴ = 16 outcomes are possible.

Let S be the set consisting of all such outcomes.

∴ n(S) = 16

Let A be the event of getting 2 tails.

∴ A = { TTHH,THTH,THHT,HTTH,HTHT,HHTT}

∴ n(A) = 6

∴ P(A) = 6/16 = 3/8

Let B be the event of getting 3 tails.

∴ B = { TTTH ,TTHT, THTT,HTTT }

⇒ n(B) = 4

∴ P(B) = 4/16 = 1/4

We need to find the probability of getting 2 tails or 3 tails i.e.

P(A ∪ B) = ?

As we can’t get 2 and 3 tails at the same time. So A and B are mutually exclusive events.

∴ P(A ∪ B) = P(A) + P(B) = 3/8 + 1/4 = 5/8

Given, Sample space is the set of first 1000 natural numbers.

∴ n(S) = 1000

Let A be the event of choosing the number such that it is multiple of 2

∴ n(A) = [1000/2] = [500] = 500 {where [.] represents Greatest integer function}

∴ P(A) =

Let B be the event of choosing the number such that it is multiple of 9

∴ n(B) = [1000/9] = [111.11] = 111 {where [.] represents Greatest integer function}

∴ P(B) =

We need to find the P(such that number chosen is multiple of 2 or 9)

∵ P(A or B) = P(A ∪ B)

Note: By definition of P(E or F) under axiomatic approach(also called addition theorem) we know that:

P(E ∪ F) = P(E) + P(F) – P(E ∩ F)

∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

We don’t have value of P(A ∩ B) which represents event of choosing a number such that number is a multiple of both 2 and 9 or we can say that it is a multiple of 18.

n(A ∩ B) = [1000/18] = [55.55] = 55

∴ P(A ∩ B) =

∴ P(A ∪ B) =

Let C represents the event that random family has colour television, W represents the event that family has black & white set and C ∩ W represents that they own both kind of sets.

According to question:

P(C) = 0.87

P(W) = 0.36 and P(C ∩ W) = 0.30

We need to find the probability of the event that a family owns either anyone kind of set.

i.e we need to find P(C or W) = P(C ∪ W) = ?

Note: By definition of P(E or F) under axiomatic approach(also called addition theorem) we know that:

P(E ∪ F) = P(E) + P(F) – P(E ∩ F)

∴ P(C ∪ W) = P(C) + P(W) – P(C ∩ W)

⇒ P(C ∪ W) = 0.87 + 0.36 – 0.30 = 0.93

Given A and B are two mutually exclusive events

And, P(A) = 0.35 P(B) = 0.45

By definition of mutually exclusive events we know that:

P(A ∪ B) = P(A) + P(B)

We have to find-

i) P(A ∪ B) = P(A) + P(B) = 0.35 + 0.45 = 0.8

ii) P(A ∩ B) = 0 {∵ nothing is common between A and B}

iii) P(A ∩ B’) = This indicates only the part which is common with A and not B ⇒ This indicates only A.

P(only A) = P(A) – P(A ∩ B)

As A and B are mutually exclusive So they don’t have any common parts ⇒ P(A ∩ B) = 0

∴ P(A ∩ B’) = P(A) = 0.35

iv) P(A’ ∩ B’) = P(A ∪ B)’ {using De Morgan’s Law}

⇒ P(A’ ∩ B’) = 1 – P(A ∪ B) = 1 – 0.8 = 0.2

Clearly according to questions sample space contains 9 elementary events(events with single outcome)

Let S represents the sample space.

∴ S = E₁∪ E₂∪ E₃ ∪ ……∪ E₈∪ E₉

Given A = {E₁, E₅, E₈}

Or A = E₁∪ E₅∪ E₈

P(E₁) = P(E₂) = 0.08, P(E₃) = P(E₄) = 0.1, P(E₆) = P(E₇) = 0.2, P(E₈) = P(E₉) = 0.07

∴ P(A) = P(E₁∪ E₅∪ E₈) = P(E₁)+P(E₅) + P(E₈)

⇒ P(A) = 0.08 + P(E₅) + 0.07 = 0.15 + P(E₅) …(1)

P(E₅) is missing, so we need to find it.

Given,

B = {E₂, E₈, E₉} or B = E₂∪ E₈∪ E₉

∴ P(B) = P(E₂∪ E₈∪ E₉) = P(E₂)+P(E₈) + P(E₉) = = 0.08+0.07+0.07 = 0.21

∴ P(B) = 0.21 ….ans(i)

∴ B’ = {E₁, E₃, E₄, E₅, E₆, E₇} or B’ = E₁∪ E₃∪ E₄∪ E₅∪ E₆∪ E₇

∴ P(B’) = P(E₁) + P(E₃) + P(E₄) + P(E₅) + P(E₆) + P(E₇)

1 – 0.21 = 0.08 + 0.1 + 0.1 + P(E₅) + 0.2 + 0.2

⇒ 0.79 = 0.68 + P(E₅)

∴ P(E₅) = 0.79 – 0.68 = 0.11

∴ from equation 1, we get-

P(A) = 0.15 + P(E₅) = 0.15 + 0.11 = 0.26 (i)

Clearly A ∩ B = {E₈}

∴ P(A ∩ B) = P(E₈) = 0.07 (i)

Using addition law of probability we know that-

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

⇒ P(A ∪ B) = 0.26 + 0.21 – 0.07

∴ P(A ∪ B) = 0.4 (ii)

As, A ∪ B = {E₁, E₅, E₈} ∪ {E₂, E₈, E₉}

⇒ A ∪ B = {E₁, E₅, E₈, E₂, E₉}

∴ P(A ∪ B) = P(E₁)+ P(E₅)+ P(E₈)+ P(E₂)+ P(E₉)

⇒ P(A ∪ B) = 0.08 + 0.11 + 0.07 + 0.08 + 0.07 = 0.41

∴ P(A ∪ B) = 0.41 (iii)

As , P(B) = 0.21

∴ P(B’) = 1 – 0.21 = 0.79 (iv)

Calculation of P(B’) using sets –

B’ = {E₁, E₃, E₄, E₅, E₆, E₇} or B’ = E₁∪ E₃∪ E₄∪ E₅∪ E₆∪ E₇

∴ P(B’) = P(E₁) + P(E₃) + P(E₄) + P(E₅) + P(E₆) + P(E₇)

P(B’) = 0.08 + 0.1 + 0.1 + 0.11 + 0.2 + 0.2

= 0.79 (iv)

Clearly through both the ways we get the same answer.

Let E denote the event that the chosen numbers are consecutive.

No of ways in which 3 numbers can be chosen out of 30 = ³⁰C₃

As we have to select 3 consecutive numbers, if we select 1 number other two are already selected.

As 29,30 can’t be selected because if they are selected we won’t be able to get 3 consecutive numbers.

∴ number of ways in which 3 consecutive numbers can be selected = number of ways in which 1 number can be chosen out of numbers from 1 to 28 = ²⁸C₁ ways

∴ P(E) =

Thus, P(E) =

Let E denote the event that the selected persons are sitting together.

As 2 persons can be selected out of n in ⁿC₂ ways

Out of n persons we can select two persons sitting together in (n-1) ways.

Because we have to select only one person next person is going to be automatically selected.

We can’t select last person because no one is sitting next to him.

∴ 1 person out of n-1 persons can be selected in (n-1) ways.

∴ P(E) =

Thus, P(E) =

As there are 9 distinct letters in “PROBABILITY” (P,R,O,B,A,I,L,T,Y).

∴ A single letter can be selected in 9 ways.

As there are 3 vowels in “PROBABILITY”.

∴ 1 vowel can be selected in 3 ways

∴ P(selected letter is vowel) = 3/9 = 1/3

In a leap year we have 366 days. So everyday of a week comes 52 times in 364 days.

Now we have 2 days remaining.

These 2 days can be-

S = {MT, TW, WTh, ThF, FSa, SaSu, SuM}

Where S is the sample space.

As there are total 7 possibilities.

∴ n(S) = 7

Let A denote the event of getting 53 Fridays and B denote event of getting 53 Saturdays.

We have to find P(A ∪ B)

Clearly,

P(A) = 2/7

P(B) = 2/7

And P(A ∩ B) = 1/7

∴ P(A ∪ B) = 2/7 + 2/7 – 1/7 = 3/7

As 3 dice are thrown. So total possible outcomes = 6×6×6 = 216

For getting 15 as the sum we need this combination-

(5, 5, 5) or (6,4,5) or (6,6,3)

(6, 4, 5) can be arranged in 3! = 6 ways.

(6, 4, 5) can be arranged in (3!)/(2!) = 3 ways

∴ 10 favourable outcomes are possible.

∴ P(E) = 10/216

Letters of MISSISSIPPI has 11 letters. Out of which 4 S’s are repeating, 4 I’s are repeating and 2 P’s are also repeating.

By Permutation theory.

Letters of MISSISSIPPI can be arranged in

Take 4 S’s together and treat them as single letter. Now we have

8 letters in together – MI(SSSS)IIPPI → MI(X)IIPPI

Letters of MI(X)IIPPI can be arranged in

∴ P(4’S come together) =

13^th day is going to be any of the day from Monday to Sunday. So probability that 13^th day of a month being Friday is 1/7.

But here the month is not declared. So the probability of selecting a random month is 1/12

By multiplication theory of Probability -

∴ probability that the 13^th days of a randomly chosen month is Friday = (1/7)×(1/12) = 1/84

3 vertices out of 6 can be chosen in ⁶C₃ = 20 ways

In a hexagon only 2 equilateral triangles are possible.

∴ probability that the triangle with these vertices is equilateral = 2/20 = 1/10

As E₁ and E₂ are independent events.

∴ P((E₁∪ E₂)∩(E₁’ ∩ E₂’))

As by De Morgan’s law.

E₁’ ∩ E₂’ = (E₁∪ E₂)’

∴ P((E₁∪ E₂)∩(E₁’ ∩ E₂’)) = P((E₁∪ E₂)∩(E₁∪ E₂)’)

⇒ P((E₁∪ E₂)∩(E₁∪ E₂)’) = P(empty set) = 0

As A and B are independent events.

∴ P(A ∩ B) = P(A)P(B) = 1/6

Also, P(A’ ∩ B’) = P(A’)P(B’) = 1/3

⇒ (1 – P(A))(1 – P(B)) = 1/3

⇒ 1 + P(A)P(B) – P(A) – P(B) = 1/3

⇒ 1 + 1/6 – 1/3 = P(A) + P(B)

⇒ 1 – 1/6 = P(A) + P(B)

∴ P(A) = 5/6 – P(B)

Let P(B) = x

∴ (5/6 – x)x = 1/6

⇒ 5x – 6x² = 1

⇒ 6x² – 5x + 1 = 0

⇒ 6x² – 3x – 2x + 1 = 0

⇒ 3x(2x – 1) – (2x – 1) = 0

⇒ (3x – 1)(2x – 1)=0

∴ x = 1/3 or x = 1/2

∴ P(B) = 1/3 or 1/2

∴ P(A) = 1/2 or 1/3

As a card is drawn from a deck of 52 cards

Let S denotes the event of card being a spade and K denote the event of card being King.

As we know that a deck of 52 cards contains 4 suits (Heart ,Diamond ,Spade and Club) each having 13 cards. The deck has 4 king cards one from each suit.

We know that probability of an event E is given as-

P(E) =

Where n(E) = numbers of elements in event set E

And n(S) = numbers of elements in sample space.

Hence,

P(S) =

P(K) =

And P(S ∩ K) =

We need to find the probability of card being spade or king, i.e.

P(Spade ‘or’ King) = P(S ∪ K)

Note: By definition of P(A or B) under axiomatic approach(also called addition theorem) we know that:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

∴ P(S ∪ K) = P(S) + P(K) - P(S ∩ K)

⇒ P(S ∪ K) =

∴ P(S ∪ K) = 4/13

∴ Option (c) is the correct choice.

If a dice is thrown twice, it has a total of (6 × 6) 36 possible outcomes.

If S represents the sample space then,

n(S) = 36

Let A represent events the event such that digit greater than 3 comes in the second throw.

∴ A = {(1,4), (2,4), (3,4), (4,4), (5,4), (6,4), (1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (1,6), (2,6), (3,6), (4,6), (5,6), (6,6)}

⇒ P(A) =

Let B represent events the event such digit greater than 3 comes in the first throw.

Similarly 18 outcomes are possible as were present for event A

⇒ P(B) =

Clearly (4,4), (5,4), (6,4), (4,5), (5,5), (6,5) (4,6), (5,6) and (6,6) are common in both events-

∴ P(A ∩ B) =

We need to find the probability of event such that at least one of the 2 throws give digit greater than 33 i.e. P(A or B) = P(A ∪ B)

Note: By definition of P(E or F) under axiomatic approach(also called addition theorem) we know that:

P(E ∪ F) = P(E) + P(F) – P(E ∩ F)

∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

⇒ P(A ∪ B) =

Hence,

P(at least one of the two throws shows digit >3) =

∴ Option (b) is correct choice.

As 2 dice are thrown so there are 6×6 = 36 possibilities.

Let E denote the event of getting a total score of 5.

E = {(1,4),(2,3),(3,2),(4,1)}

∴ n(E)=4

Hence,

P(E) = 4/36 = 1/9

As our answer matches only with option (c)

∴ Option (c) is the only correct choice.

As 2 dice are thrown so there are 6×6 = 36 possibilities.

Let E denote the event of getting a total score of 5.

E = {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}

∴ n(E)= 6

Hence,

P(E) = 6/36

As our answer matches only with option (b)

∴ Option (b) is the only correct choice.

As 2 dice are thrown so there are 6×6 = 36 possibilities.

Let E denote the event of getting a total score of 5.

E = {(4,6),(5,5),(6,4)}

∴ n(E)= 3

Hence,

P(E) = 3/36 = 1/12

As our answer matches only with option (b)

∴ Option (b) is the only correct choice.

As a card is to be drawn from 1 to 100.

∴ n(S) = 100

As we know that largest no less that or equal to 100 which is a perfect square = 100

And, √100 = 10

∴ there are total 10 numbers from 1 to 100 which are square.

∴ n(E) = 10

Hence,

P(E) = 10/100 = 1/10

As our answer matches only with option (c)

∴ Option (c) is the only correct choice.

As there are 3 +4+5 = 12 balls.

So, 2 balls out of 12 can be drawn in ¹²C₂ ways = 66

Let E denote the event that balls drawn are of different colours

Either the balls will be red-white or white-blue or red-blue.

∴ P(E) =

Hence,

P(E) = 47/66

As our answer matches only with option (a)

∴ Option (a) is the only correct choice.

In a single throw of 2 die, we have total 36(6 × 6) outcomes possible.

Say, n(S) = 36 where S represents Sample space

Let A denotes the event of getting a doublet(equal number)

∴ A = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

∴ P(A) =

And B denotes the event of getting a total of 9

∴ B = {(3,6), (6,3), (4,5), (5,4)}

P(B) =

We need to find probability of the event of getting neither a doublet nor a total of 9.

P(A’ ∩ B’) = ?

As, P(A’ ∩ B’) = P(A ∪ B)’ {using De Morgan’s theorem}

P(A’ ∩ B’) = 1 – P(A ∪ B)

Note: By definition of P(E or F) under axiomatic approach(also called addition theorem) we know that:

P(E ∪ F) = P(E) + P(F) – P(E ∩ F)

∴ P(A ∪ B) = {As P(A ∩ B) = 0 since nothing is common in set A and B ⇒ n(A ∩ B) = 0 }

Hence,

P(A’ ∩ B’) = 1 – (5/18) = 13/18

Hence,

P(required event) = 13/18

As our answer matches only with option (b)

∴ Option (b) is the only correct choice.

As there are 3 +2+4 = 9 persons

So, 4 persons out of 9 can be drawn in ⁹C₄ ways = 126

Let E denote the event that there are exactly 2 children in the selection.

2children out of 4 can be selected in ⁴C₂ = 6 ways

And rest two persons can be male or female. So we will select 2 persons out of remaining 5.

∴ P(E) =

Hence,

P(E) = 10/21

As our answer matches only with option (c)

∴ Option (c) is the only correct choice.

Given, P(A) = 0.25 and P(B) = 0.5

Also P(A ∩ B) = 0.14

We have to find P(A’ ∩ B’)

By De Morgan’s theorem we know that:

P(A’ ∩ B’) = P(A ∪ B)’

We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

∴ P(A ∪ B) = 0.25 + 0.5 – 0.14 = 0.61

∴ P(A’ ∩ B’) = P(A ∪ B)’ = 1 -P(A ∪ B) = 1 – 0.61 = 0.39

As our answer does not match with any other option

But option of none of these is there. So most suitable option is option (d)

∴ Option (d) is the only correct choice.

When a die is rolled there are in total 6 possibilities.

∴ n(S) = 6

Let E denote event of getting 1 or 6

∴ n(E) = 2

∴ P(E) = 2/6 = 1/3

Hence,

P(E) = 1/3

As our answer matches only with option (c)

∴ Option (c) is the only correct choice.

As 6 boys and 6 girls are sitting in a row. So these 12 persons can sit in 12! Ways

Now group all 6 girls together and treat them as 1.

Now, all girls together can sit in 7! Ways

And girls can sit among self in 6! Ways.

∴ total ways in which all 6 girls sit together = 7! × 6!

∴ P(E) =

Hence,

P(E) = 1/132

As our answer matches only with option (d)

∴ Option (d) is the only correct choice.

As we know that 3 events are said to be mutually exclusive events if –

P(A ∪ B ∪ C) = P(A) + P(B) + P(C)

⇒ P(A ∪ B ∪ C) =

As probability can never be > 1

∴ A,B and C are not mutually exclusive events.

Statement is false

∴ Option (b) is the only correct choice.

As 3 events are mutually exclusive and exhaustive.

∴ P(A ∪ B ∪ C) = P(A) + P(B) + P(C) = 1

⇒

⇒

⇒ 5 + 3p = 6

∴ p = 1/3

As 1/3 lies in both intervals given in option (a) and (d)

But a larger value like 100 which comes if we choose option d can make P(A) < 0 which is not possible.

∴ Option (a) is the most suitable choice here.

As total number of cards = 16

2 cards can be drawn in ¹⁶C₂ ways = 120 ways

As we need to find the probability for the event E such that at least one of them is an ace.

We can solve this problem using negation.

We will find the probability for event that both cards are ace (E’)

Automatically 1 – P(E’) will give P(E)

For finding P(E’) we will select both cards out of 4 ace cards

∴ P(E’) =

∴ P(E) = 1 – 1/20 = 19/20

Hence,

P(E) = 19/20

As our answer matches only with option (c)

∴ Option (c) is the only correct choice.

As 3 dice are thrown so there are 6×6×6 = 216 possibilities.

Let E denote the event of getting a total score of 5.

{(1,1,3),(1,2,2)}

As (1,1,3) can be arranged in

and (1,2,2) can be arranged in

∴ n(E) = 6

Hence,

P(E) = 6/36 = 1/6

As our answer matches only with option (b)

∴ Option (b) is the only correct choice.

Let E and F be the two events such that one must occur.

Given,

P(E) = 2/3 P(F)

Also, P(E ∪ F) = 1

P(E) + P(F) = 1

⇒ P(F){2/3 + 1} = 1

∴ P(F) = 3/5

And P(F’) = 1 – 3/5 = 2/5

We have to find

∴ Odds in favour of F = 3/2

As our answer matches only with option (d)

∴ Option (d) is the only correct choice.

In a leap year we have 366 days. So, every day of a week comes 52 times in 364 days.

Now we have 2 days remaining.

These 2 days can be-

S = {MT, TW, WTh, ThF, FSa, SaSu, SuM}

Where S is the sample space.

As there are total 7 possibilities.

∴ n(S) = 7

Let A denote the event of getting 53 Fridays and B denote event of getting 53 Saturdays.

We have to find P(A ∪ B)

Clearly,

P(A) = 2/7

P(B) = 2/7

And P(A ∩ B) = 1/7

∴ P(A ∪ B) = 2/7 + 2/7 – 1/7 = 3/7

As our answer matches only with option (b)

∴ Option (b) is the only correct choice.

As we have 4 letters and 4 envelopes.

These 4 letters can be arranged in 4! = 24 ways.

∴ n(S) = 24

Let E denotes the event that all letters are not placed in the right envelopes

The number of ways in which 4 letters can be placed in wrong envelopes is given by the number of ways in which N objects can be dearranged.

Numbers of ways of in which N objects can be dearranged is given by –

∴ n(E) =

∴ P(E) = 9/24 = 3/8

Given, P(A) = 0.25 and P(B) = 0.5

Also P(A ∩ B) = 0.14

We have to find P(A’ ∩ B’)

By De Morgan’s theorem we know that:

P(A’ ∩ B’) = P(A ∪ B)’

We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

∴ P(A ∪ B) = 0.25 + 0.5 – 0.14 = 0.61

∴ P(A’ ∩ B’) = P(A ∪ B)’ = 1 -P(A ∪ B) = 1 – 0.61 = 0.39

As our answer matches with option (a)

∴ Option (a) is the only correct choice.

Let A be the event that student ‘A’ fails in exam and B be the event that student ‘B’ fails in exam.

Given, P(A) = 1/5 and P(B) = 3/10

We have to find P(A ∪ B)

We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

As P(A ∩ B) not given, so assuming its value 0

∴ P(A ∪ B) = 1/5 + 3/10 = 5/10 = 1/2

As our answer matches with option (a)

∴ Option (a) is the only correct choice.

As total articles = 10 + 6 = 16

P(good) =P(G) = 10/16

And P(defective)= P(D) = 6/16

As there it is not possible to select an item that is both good and defective.

P(good or Defective) = P(G ∪ D) = P(G) + P(D) = 16/16 = 1

As our answer does not directly seems to match with any option.

But option (a) is 64/64 = 1

∴ option (a) is the only correct choice.

3 integers out of 20 can be chosen in ²⁰C₃ ways.

As in first 20 integers, 10 are even.

To have the product of 3 integer even the chosen integers must be even.

∴ 3 integers that are even can be chosen from first 20 integers in ¹⁰C₃ ways.

∴ Probability =

Our answer matches with option (a)

∴ Option (a) is the only correct choice

2 integers out of 30 can be chosen in ³⁰C₂ ways.

As in first 30 integers, 15 are even and 15 are odd.

To have the sum of 2 integer odd one integer must be even and other must be odd.

∴ integers can be chosen in ¹⁵C₁ × ¹⁵C₁ ways

∴ Probability =

Our answer matches with option (c)

∴ Option (c) is the only correct choice

According to question bag contains 5+4+3 = 12 balls

∴ 1 ball can be chosen in ¹²C₁ = 12 ways

Let B denotes the event of drawing a black ball and R denotes event of drawing red ball

∴ P(B) = 5/12 and P(R) = 3/12

We have to find P(B ∩ R)

As, we need to find P(B ∪ R)

∵ Black and red balls can’t be drawn simultaneously.

∴ P(B ∩ R) = 0

∴ P(B ∪ R) = P(B) + P(R) = 5/12 + 3/12 = 8/12 = 2/3

Our answer matches with option (d)

∴ Option (d) is the only correct choice

Sample Space = 36

Out of 36 possible events, only one event is favourable = (6, 6)

Probability

Balls in urn = 9

3 balls can be drawn in ⁹C₃ ways.

To draw the balls of same colour, we can select balls all three red, all blue or all black.

P(ball of same colour) = P(all blue) or P(all black) or P(all red)

P(all red) = not possible as we have to select 3 balls but there are only 2 red balls.

∴ P(All red) = 0

P(all black) =

And P(all blue) =

∴ P(ball of same colour) = P(all blue) or P(all black) or P(all red)

⇒ P(ball of same colour) = P(all blue) + P(all black) + P(all red)

= 1/84 + 1/21 + 0 = 5/84

Our answer matches with option (a)

∴ Option (a) is the only correct choice

As building has 8 floors including ground.

So, they have 7 options to leave the lift.

∴ total ways in which persons can leave the lift = 7⁵

As the number of ways in which all persons leave in a different floor can be given by: 7× 6× 5× 4× 3 = ⁷P₅

∴ P(that all leave lift in different floor) =

Our answer matches with option (a)

∴ Option (a) is the only correct choice

As total articles = 10 + 6 = 16

P(good) =P(G) = 10/16

And P(defective)= P(D) = 6/16

As there it is not possible to select an item that is both good and defective.

P(good or Defective) = P(G ∪ D) = P(G) + P(D) = 16/16 = 1

As our answer does not directly seems to match with any option.

But option (a) is 64/64 = 1

∴ option (a) is the only correct choice.

As total articles(nuts + nails) = 10 + 6 = 16

Number of rusted nails = 3

Number of rusted nuts = 5

Number of rusted articles = 8

Let N denotes the event that the article drawn is a nails and R be the event denoting drawn article is rusted

P(N) = 6/16

And P(R) = 8/16

Also, we have possibility that drawn nail is rusted.

∴ P(N ∩ R) = 3/16

P(Nail or Rusted) = P(N ∪ R) = P(N) + P(R) – P(N ∩ R) = 6/16 + 8/16 – 3/16 = 11/16

∴ option (c) is the only correct choice.

As , S = A ∪ B

∴ P(A ∪ B) = P(S) = 1

As A and B are mutually exclusive events.

∴ P(A ∪ B) = P(A) + P(B)

⇒ 1 = P(A) + 3P(A) [∵ P(A) = 1/3 P(B)]

⇒ 4P(A) = 1

∴ P(A) = 1/4

Our answer matches with option (a)

∴ Option (a) is the only correct choice

We know that the no of mapping from a set A to same set containing n elements = nⁿ

As each element have n options to be mapped with n elements.

For mapping selected to be one to one, for first element we have n options ,for second we have n-1 options an so on…

∴ total such mappings = n × (n-1)× (n-2)× … = n!

∴ P(mapping is one to one) =

Our answer matches with option (c)

∴ Option (c) is the only correct choice

Given,

3P(A) = 2P(B) = P(C) = k(say)

∴ P(A) = k/3

P(B) = k/2

And P(C) = k

As events A,B and C are mutually exclusive and exhaustive,

∴ P(A ∪ B ∪ C) = P(A) + P(B) + P(C) = 1

⇒ 1 = k/3 + k/2 + k

⇒ 1 =

⇒ k = 6/11

As, P(A) = k/3 =

Our answer matches with option (b)

∴ Option (b) is the only correct choice

As P(A) and P(B) are mutually exclusive event.

∴ P(A ∪ B) = P(A) + P(B)

As P(A ∪ B) ≤ 1

∴ P(A) + P(B) ≤ 1

⇒ P(A) ≤ 1 – P(B)

⇒ P(A) ≤ P(B’)

Our answer matches with option (a)

∴ Option (a) is the only correct choice

As, (A ∪ B) = (A ∩ B)

⇒ A and B are same sets.

∴ P(A) = P(B)

Our answer matches with option (a)

∴ Option (a) is the only correct choice

3 numbers can be chosen out of 20 in ²⁰C₃ ways.

As we have to choose 3 consecutive numbers. So we can’t choose 19 and 18.

Choosing consecutive number is same as choosing a single number because other 2 numbers are automatically chosen.

This can be chosen 18 ways.

∴ P(chosen numbers are consecutive) =

Our answer matches with option (d)

∴ Option (d) is the only correct choice

As 6 boys and 6 girls are sitting in a row. So these 12 persons can sit in 12! Ways

Now group all 6 girls together and treat them as 1.

Now, all girls together can sit in 7! Ways

And girls can sit among self in 6! Ways.

∴ total ways in which all 6 girls sit together = 7! × 6!

∴ P(E) =

Hence,

P(E) = 1/132

As our answer matches only with option (d)

∴ Option (d) is the only correct choice.

Total numbers of 4digit that can be formed using 0,2,3,5 without repetition= 3× 3 × 2 × 1 = 18

For number to be divisible by 5 unit digit must end with 5 or 0.

∴ total such numbers(4 digit) ending with zero are = 3× 2× 1(0 is fixed at the unit digit) = 6

∴ total such numbers(4 digit) ending with zero are = 2× 2× 1(5 is fixed at the unit digit) = 4

Total 4 digit numbers using given digit divisible by 5 are 6 + 4 =10

∴ P(a number divisible by 5) = 10/18 = 5/9

As our answer matches only with option (d)

∴ Option (d) is the only correct choice.

Let A be the event that student ‘A’ fails in exam and B be the event that student ‘B’ fails in exam.

Given, P(A) = 0.2 and P(B) = 0.3

We have to find P(A ∪ B)

We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

As P(A ∩ B) not given, but we know that – 0 ≤ P(A ∩ B) ≤ 1

∴ P(A ∪ B) = 0.2 + 0.3 – P(A ∩ B) = 0.5 - P(A ∩ B)

Clearly a +ve number will be subtracted form 0.5 to make it equal to P(A ∪ B).

∴ P(A ∪ B) ≤ 0.5

As our answer matches with option (c)

∴ Option (c) is the only correct choice.

Total numbers of 3 digit that can be formed using 0,2,4,6,8 = 4× 5 × 5 = 100

3-digit number formed using 0,2,4,6,8 with all digits same are 222,444,666,888

∴ P(number has the same digits) = 4/100 = 1/25

As our answer matches only with option (d)

∴ Option (d) is the only correct choice.