Probability

Given that for a loaded die-

P (1) = P (2) = 0.2, P (3) = P (5) = P (6) = 0.1 and P (4) = 0.3

And given that die is thrown two times and A is the event of same number each time

B is the event of a total score is 10 or more.

So,

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

P(A)= [P (1,1) +P (2,2) +P (3,3) + P (4,4) + P (5,5) + P (6,6)]

P(A)= [P (1) × P (1) + P (2) × P (2) + P (3) × P (3) + P (4) × P (4) +P (5) × P (5) + P (6) × P (6)]

P(A)= [0.2×0.2+ 0.2×0.2+ 0.1×0.1+0.3×0.3+ 0.1×0.1+ 0.1

×0.1]

P(A)= [0.04+ 0.04+ 0.01+ 0.09+ 0.01+0.01]

P(A)= [0.20]

&

B= EVENT OF TOTAL SCORE IS 10 OR MORE

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

P(B)= [P (4,6) + P (5,5) + P (5,6) + P (6,4) + P (6,5) + P (6,6)]

P(B)= [P (4) ×P (6) + P (5) ×P (5) + P (5) ×P (6) + P (6) ×P (4) + P (6) ×P (5) + P (6) ×P (6)]

P(B)= [0.3×0.1+ 0.1×0.1+ 0.1×0.1+ 0.1×0.3+ 0.1×0.1+ 0.1×0.1]

P(B)= [0.03+ 0.01+ 0.01+ 0.03+ 0.01+ 0.01]

P(B)= [0.10]

ALSO, probability of an intersection B (i.e. both the events occur simultaneously)

A Ո B = {(5,5), (6,6)}

HENCE,

P (A Ո B) = P (5,5) + P (6,6)

P (A Ո B) = P (5) × P (5) + P (6) ×P (6)

P (A Ո B) = 0.1×0.1+ 0.1×0.1

P (A Ո B) = 0.01+0.01

P (A Ո B) = 0.02

We know that if two events are independent than

P (A Ո B) = P(A) P(B)

HERE,

P(A). P(B)= 0.20× 0.10 = 0.02

SO, P (A Ո B) = P(A) P(B)

Hence, A and B are independent events.

Given-

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

So, n(A)= 6, n(S)= (6)²= 36

∴

And B = {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)}

n(B)= 6

∴

AՈB = [(5,5), (6,6)]

∴

Thus,

P (A Ո B) ≠ P(A). P(B)

So, A and B are not independent events.

Given that

probability that at least one of the two events A and B occurs is 0.6 i.e. P(AՍB) = 0.6

& The probability that A and B occur simultaneously is 0.3 i.e. P(AՈB) = 0.3

We know that:

P(AՍB) = P(A)+ P(B) – P(AՈB)

0.6 = P(A)+ P(B) – 0.3

P(A)+ P(B) = 0.6+ 0.3 = 0.9

We have to find

So,

Given that bag contains 5 red marbles and 3 black marbles

For at least one of the three marbles drawn be black, if the first marble is red

Then the following three conditions will be followed

(i) Second marble is black and third is red = E₁

(ii) Second and third, both marbles are black = E₂

(iii) Second marble is red and third marble is black = E₃

Let event R_n = drawing red marble in n^th draw

And event B_n = drawing black marble in n^th draw

And

Required probability, P(E)= P(E₁) + P(E₂) + P(E₃)

Given that two dice are drawn together i.e. n(S)= 36

Where S is sample space

Also given that

E = a of total 4

∴E = {(2,2), (3,1), (1,3)}

∴n(E) = 3

And F= a total of 9 or more

∴ F = {(3,6), (6,3), (4,5), (5,4), (6,4), (4,6), (6,5), (6,6), (5,5), (5,6)}

∴n(F)=10

And,

G = a total divisible by 5

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

∴ n(G) = 7

Here, (E Ո F) = φ AND (E Ո G) = φ

Also, (F Ո G) = {(4,6), (6,4), (5,5)}

n (F Ո G) = 3 and (E Ո F Ո G) = φ

And,

So,

P (F Ո G) ≠ P(F). P(G)

hence, there is no pair which is independent

Let p and q denote the events of failure and success, respectively.

We know that, a random variable X (=0,1, 2,…., n) is said to have Binomial parameters n and p, if its probability distribution is given by

Where, q = 1-p

And r =0,1,2,…..n

In an experiment of tossing a coin three times, we have n=3 and random variable X can take values r =0,1,2 and 3 with and

So, we see that in the experiment of tossing a coin three times, we have random variable X which can take values 0,1,2 and 3 with parameters n=3 ad

Hence tossing of a coin 3 times is a Binomial distribution.

Given,

(i)

(ii)

(iii)

(iv)

GIVEN THAT

AND,

AND

By De Morgan’s laws:

P (A'∩ C') = P(A Ս C)^’

= 1- P (A Ս C)

= 1-[P(A)+ P(C)- P (A Ո C)]

Given that P(E₁) =P₁ and P(E₂) =P₂

(i)P₁P₂

=P(E₁). P(E₂) =P (E₁ Ո E₂)

So, E₁ and E₂ occur simultaneously.

(ii)(1-P₁) P₂

As we know P(A)+P(A)^’ = 1

=P(E₁)’. P(E₂)

= P (E₁’ Ո E₂)

So, E₁ does not occur but E₂ occur.

(iii)1-(1-P₁) (1-P₂)

As we know P(A)+P(A)^’ = 1

=1-P(E₁)’P(E₂)’

=1-P (E₁’ Ո E₂’)

By De Morgan’s laws:

=1-P (E₁ Ս E₂)^’

As we know P(A)+P(A)^’ = 1

= 1- [1-P (E₁ Ս E₂)]

= P (E₁ Ս E₂)

So, either E₁ or E₂ or both E₁ and E₂ occurs.

(iv)P₁+P₂-2P₁P₂

= P(E₁) + P(E₂) – 2P(E₁) P(E₂)

= P(E₁) + P(E₂) – 2P (E₁ Ո E₂)

= P (E₁ Ս E₂)- 2P (E₁ Ո E₂)

So, either E₁ or E₂ occurs but not both.

Given table—

(i) We know that , where pi ≥0

P₁+ P₂+ P₃+ P₄ = 1

k+ k²+2k²+ k = 1

3k² + 2k -1= 0

3k² + 3k- k- 1 = 0

3k(k+1) -1 (k+1) = 0

Since, k≥ 0, we take

(ii)

= 0.5(k)+ 1(k²) + 1.5(2k²) + 2(k)

= 4k² + 2.5k

(i)

WE HAVE TO PROVE,

P(A)= P(AՈB) + P(AՈ)

We know A = AՈS

We can write s = A Ս A^’ and s = B Ս B^’

So,

A = A Ո (B Ս B^’)

= (A Ո B) Ս (A Ո B^’)

Two events are mutually exclusive or disjoint if they cannot both occur at the same time.

(A Ո B) means A and B both occurring at the same time while (A Ո B^’) means A and B^’ both occurring at the same time.

So, it is not possible that (A Ո B) and (A Ո B^’) occur at the same time.

Hence (A Ո B) and (A Ո B^’) are mutually exclusive.

When events are mutually exclusive then P (A Ո B) = 0

∴ P[(A Ո B) Ո (A Ո B^’)] = 0 ….. (1)

So, A = A Ո (B Ս B^’)

As we know P (A Ս B) = P (A) + P(B) - P (A Ո B)

P(A) = P [(A Ո B) Ս (A Ո B^’)]

= P (A Ո B) + P (A Ո B^’) – P [(A Ո B) Ո (A Ո B^’)]

From (1),

P(A) = P(AՈB) + P(AՈ)

Hence proved

(ii)WE HAVE TO PROVE, P(AՍB) = P(AՈB) + P(AՈ) + P(ՈB)

AՍB means the all the possible outcomes of both A and B.

From the Venn diagram we can see,

AՍB = (AՈB) Ս (AՈ) Ս (ՈB)

Two events are mutually exclusive or disjoint if they cannot both occur at the same time.

(A Ո B) means A and B both occurring at the same time while (A Ո B^’) means A and B^’ both occurring at the same time.

(ՈB) means A^’ and B both occurring at the same time.

So, it is not possible that (A Ո B), (A Ո B^’) and (ՈB) occur at the same time.

Hence (A Ո B), (A Ո B^’) and (ՈB) are mutually exclusive.

When events are mutually exclusive then P (A Ո B) = 0

P [(A Ո B) Ո (A Ո B^’)] =0 ….. (1)

P [(A Ո B^’) Ո P(ՈB)] = 0 ….. (2)

P [(A Ո B) Ո P(ՈB)] = 0 ….. (3)

P [(A Ո B) Ո (A Ո B^’) Ո P(ՈB)] = 0 …. (4)

P(AՍB) = P[(AՈB) Ս (AՈ) Ս (ՈB)]

We know,

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) − P(B ∩ C) + P(A ∩ B ∩ C)

So,

P(AՍB) = P[(AՈB) Ս (AՈ) Ս (ՈB)] = P(AՈB) + P(AՈ) + P(ՈB) – P[(AՈB) ∩ (AՈ] – P[(AՈB) ∩ (ՈB)]− P[(AՈ) ∩ (ՈB)] + P [(A Ո B) Ո (A Ո B^’) Ո P(ՈB)]

From (1), (2), (3) and(4) we get,

P(AՍB) = P(AՈB) + P(AՈ) + P(ՈB)

Hence proved.

Given that, Radom variable X is the member of tails in three tosses of a coin

So, X= 0,1,2,3

Where n=3, and x= 0,1,2,3

We know that, Var (X)= E(X²)- [E(X)]² ……(i)

Where, E(X²) =

=3

And

Putting the values of E(X)² and [E(X)]² in equation (i), we get:

And standard deviation of

Let X is the random variable of profit per throw.

Probability of getting any number on dice is .

Since, she loss Rs 1 on getting any of 2, 4 or 5.

So, at X= -1,

P(X) = P (2) +P(4) +P(5)

Similarly, =1 if dice shows of either 1 or 6.

P(X) = P (1) +P (6)

and at X=4 if die shows a 3

P(X) = P (3)

∴ Player’s expected profit = E(X)= XP(X)

As three dice are thrown at the same time, so we have sample space [n(S)] = 6³= 216

Let E₁ is the event when the sum of numbers on the dice was six and E₂ is the event when three twos occurs.

E₁= {(1,1,4), (1,2,3), (1,3,2), (1,4,1), (2,1,3), (2,2,2,), (2,3,1), (3,1,2), (3,2,1), (4,1,1)}

n(E₁) =10 and E₂ {2,2,2}

n(E₂) =1

Also, (E₁ Ո E₂) = 1

Let X variable for the prize.

There is possibility of winning nothing, Rs 500,Rs 2000 and Rs 3000.

So, X will take these values.

Since there are 3 third prizes of 500 probability of winning third prize is .

1 first prize of 3000 probability of winning third prize is .

1 second prize of 2000 probability of winning third prize is .

Since, E(X)= X(PX)

Given that-

W₁= [4 white balls] and B₁ = [5 black balls]

And W₂= [9 white balls] and B₂= [7 black balls]

Let E₁ is event that the ball transferred from the first bag is white and E₂ is the event that the ball transferred from the bag is black.

Also, E is the event that the ball drawn from the second bag is white.

And

P(E)=P (E₁). P (E|E₁) +P (E₂). P(E|E₂)

Given that

Bag I= [3Black, 2White], Bag II= [2 black, 4 white]

Let E₁= event that bag I is selected

E₂= event that bag II is selected

E₃= event that a black ball is selected

∴P(E)= P(E₁). P(E|E₁) +P(E₂). P(E|E₂)

Given that box has 5 blue and 4 red balls.

Let E₁ is the event that first ball drawn is blue, E₂ is the event that the first ball drawn is red and E is the event that second ball drawn is blue.

∴P(E)= P(E₁). P(E|E₁) + P(E₂). P(E|E₂)

Let E₁, E₂, E₃ and E₄ are the events that the first, second, third and fourth card is king respectively.

As there are 4 kings,

when 1 king is taken out kings left are 3 and total cards will be 51.

So, probability of drawing a king when one king has been taken out is:

Now when 2 kings taken out 2 kings are left, and 50 cards are there.

So, probability of drawing a king when two kings have been taken out is:

Now when 3 kings taken out 1 king is left, and 49 cards are there.

So, probability of drawing a king when three kings have been taken out is:

Probability that all 4 cards are king is:

∴P (E₁ Ո E₂ Ո E₃ Ո E₄) = p(E₁). P(E₂|E₁). P (E₃|E₁ Ո E2). P [E₄|(E₁ ՈE₂ Ո E₃ ՈE₄)]

=

Given that,

n=5,

Odd numbers = 1,3,5

and

also, r=3

Let X be the random variable for getting a head.

Here, n=10, r≥8

r=8,9,10

We know that,

∴P(X=r) = P(r=8) + P(r=9)÷P(r=10)

Given that man shoots 7 times, so n=7

And probability of hitting target

Where,

∴P(X=r≥2) =1-[p(r=0) +P(r=1)]

given that 10 defective watches in 100 watches

probability of defective watch from a lot of 100 watch

∴

We know

We have

Var(X)=E(X²)-[E(X)]²

Where,

And

∴0+0.25+0.6+0.6+0.60= 2.05

And E(X)²=0+0.25+1.2+1.8+2.40=5.65

(i)

As we know Var(ax) = a² var(x)

(ii) V(X)

Var(X)=E(X²)-[E(X)]²

= 5.65 – (2.05)²

= 5.65 - 4.2025

= 1.4475

We have

(i) Since,

8k+ 4k+ 2k+ k= 8

(ii)

(iii)

We have,

We know that standard deviation of

Where, Var =E(X²)-[E(X)]²

∴VarX= [0.8+ 4.5+ 4.8]- [0.5+1.5+1.2]²

=10.1-(3.1)²

=10.1-9.61

=0.49

∴standard deviation of

Since, X= number of four seen

On tossing to die, X=0,1,2

Also, and

So,

Thus, we get following table 1207110331

∴Var(X)=E(X)²-[E(X)²] =X²P(x)-[XP(X)]²

Given that, X= no. of twos seen

So, on throwing a die three times, we will have X=0,1,2,3

P(X=1) = P_{not 2.} P_{not 2}. P_{2 +} P _2. P_2. P_{not 2} + P ₂. P_{not 2.} P ₂

P(X=2) =P_{not 2.} P_2. P_{2 +} P_{not 2.} P_2. P_{2 +} P₂. P_{not 2}.P₂

We know that,

Given-

For first die, and

[∵ P(1)=P(2)=P(3)=P(4)=P(5)]

For second die

Let X= number of one’s seen

For X=0,

=0.04

Hence, the required probability distribution is as below

Since, we have to prove that, E(Y²) =2E(X) -----(i)

Taking LHS of equation (i), we have:

E(Y)²= Y²P(Y)

=

……(ii)

Now taking RHS of equation (i) we get:

E(X)= XP(X)

……..(iii)

Thus, from equations (ii) and (iii), we get:

E(Y²) =2E(X)

Hence proved.

Let X is the random variable which denotes that a bulb is defective.

Also, n =10, and

(i)None of the bulbs is defective i.e., r=0

(ii)Exactly two bulbs are defective i.e., r=2

(iii)More than 8 bulbs work properly i.e., there is less than 2 bulbs which are defective.

So, r<2 r=0,1

∴P(X=r) =P(r<2) =P (0) +P (1)

=+

Let E₁= event that fair coin is drawn

E₂ = event that two headed coin is drawn

E= event that tossed coin get a head

∴

Using Bayes’ theorem, we have

Given condition can be represented as below:

Let E₁ = event that the person selected is of group O

E₂= event that the person selected is of other than blood group O

And E₃= event that selected person is left handed

∴P(E₁) =0.30, P(E₂) =0.70

P(E₃|E₁) = 0.60 And P(E₃|E₂) =0.10

Using baye’s theorem, we have:

The notation P [r ≤ p| s ≤ p]
means
P (r ≤p) *given that s ≤ p
Since we're told s ≤ p , then it means s is drawn first.
Let we have n numbers before s is drawn:
(1 . . s …. . p . . .. n)
After s is drawn,
[ 1 ... p] has one element missing, so there are (p-1) elements.
Also, the entire set has one element missing, so there are (n-1) altogether.
P(r is among p – 1 elements )

Among (1 . . s …. . p) the probability of drawing s is .
Now,

P [r ≤ p|s ≤ p] is the probability that r ≤ p when s ≤ p.

So,

Let X is the random variable score obtained when a die is thrown twice.

∴ X= 1,2,3,4,5,6

Here,

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

Similarly,

So, the required distribution is,

Also, we know that,

Given that-

X=0,1,2 and P(X) at X=0 and 1,

Let at X=2, P(X) is x.

p+p+x=1

x=1-2p

We get the following distribution

∴E(X)= XP(X)

= 0×P + 1×P+ 2(1-2P)

= P+2-4P = 2-3P

And E(X)²= X² P(X)

= 0×P+ 1× P+ 4×(1-2P)

= P+4-8P=4-7P

Also, given that E(X²) =E(X)

4-7p= 2-3p

4p=2

We have,

∴ Variance= E(X)² – [E(X)]² = X²P(X)- [XP(X)]²

According to question, A and B throw a pair of dice alternately.

A wins if he gets a total of 6

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

And B wins if she gets a total of 7

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

Let P(B) is the probability, if A wins in a throw

And P(B) is the probability, if B wins in a throw

∴probability of winning the game by A in third throw

We Have, A= {(x, y):x+y=11}

And B= {(x, y): x≠5}

∴ A = {(5,6), (6,5)}

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

n(A)=2, n(B)= 30, n(AՈB) = 1

∴

So, A and B are not independent.

Given that an urn contains m white and n black balls.

Let E₁ = first ball drawn of white colour

E₂= first ball drawn of black colour

And E₃= second ball drawn of white colour

∴

Also,

Using the probability theorem, we have:

∴P(E₃) = P(E₁). P () +P(E₂).

Hence, the probability of drawing a white ball does not depend on k.

Let E₁, E₂, and E₃ be the events that Bag 1, Bag 2 and Bag 3 is selected, and a ball is chosen from it.

Bag 1: 3 red balls,

Bag 2: 2 red balls and 1 white ball

Bag 3: 3 white balls.

As The probability that bag i will be chosen and a ball is selected from it is i|6.

The Law of Total Probability:

In a sample space S, let E₁,E₂,E₃…….E_n be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with E₁,E₂,E₃…….E_n, then

P(A) = P(E₁)P(A/E₁)+ P(E₂)P(A/E₂)+ …… P(E_n)P(A/E_n)

(i) Let “E” be the event that a red ball is selected.

P(E|E₁) is the probability that red ball is chosen from the bag 1.

P(E|E₂) is the probability that red ball is chosen from the bag 2.

P(E|E₃) is the probability that red ball is chosen from the bag 3.

So,

As red ball can be selected from Bag 1, Bag 2 and Bag 3.

So, probability of choosing a red ball is the sum of individual probabilities of choosing the red from the given bags.

From the law of total probability,

P(E) = P(E₁) × P(E|E₁) + P(E₂) × P(E|E₂) + P(E₃) × P(E|E₃)

(ii) Let F be the event that a white ball is selected.

So, P(F|E₁) is the probability that white ball is chosen from the bag 1.

P(F|E₂) is the probability that white ball is chosen from the bag 2.

P(F|E₃) is the probability that white ball is chosen from the bag 2.

P(F|E₁) = 0

As white ball can be selected from Bag 1, Bag 2 and Bag 3.

So, probability of choosing a white ball is the sum of individual probabilities of choosing the red from the given bags.

P(F) = P(E₁) × P(F|E₁) + P(E₂) × P(F|E₂) + P(E₃) × P(F|E₃)

Referring to the previous solution, using Bayes theorem, we have

Let E₁, E₂, and E₃ be the events that Bag 1, Bag 2 and Bag 3 is selected, and a ball is chosen from it.

Bag 1: 3 red balls,

Bag 2: 2 red balls and 1 white ball

Bag 3: 3 white balls.

As The probability that bag i will be chosen and a ball is selected from it is i|6.

Let F be the event that a white ball is selected.

So, P(F|E₁) is the probability that white ball is chosen from the bag 1.

P(F|E₂) is the probability that white ball is chosen from the bag 2.

P(F|E₃) is the probability that white ball is chosen from the bag 2.

P(F|E₁) = 0

We have to find the probability that if white ball is selected it is selected from:

(i) Bag 2

We use Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.

P(E₂|F) is the probability that white ball is selected from bag 2.

(ii)Bag 3

We use Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.

P(E₃|F) is the probability that white ball is selected from bag 2.

Using Bayes’ theorem, we get the probability of P(E₃|F) as:

Here, A₁, A₂, and A₃ denote the three types of flower seeds.

and A₁: A₂: A₃ = 4: 4 : 2

Total outcomes = 10

So,

Let E be the event that a seed germinates and E’ be the event that

a seed does not germinate.

Now P(E|A₁) is the probability that seed germinates when it is seed A₁.

P(E^’|A₁) is the probability that seed will not germinate when it is seed A₁.

P(E|A₂) is the probability that seed germinates when it is seed A₂.

P(E’|A₂) is the probability that seed will not germinate when it is seed A₂.

P(E|A₃) is the probability that seed germinates when it is seed A₃.

P(E’|A₃) is the probability that seed will not germinate when it is seed A₃.

and

The Law of Total Probability:

In a sample space S, let E₁, E₂, E₃……. En be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with E₁, E₂, E₃……. E_n, then

P(A) = P(E₁) P(A|E₁) + P(E₂) P(A|E₂) + …… P(E_n)P(A|E_n)

(i) Probability of a randomly chosen seed to germinate.

It can be either seed A, B or C.

So,

From law of total probability,

P(E) = P(A₁) × P(E|A₁) + P(A₂) × P(E|A₂) + P(A₃) × P(E|A₃)

= 0.49

(ii) that it will not germinate given that the seed is of type A₃

As we know P(A) + P(A^’) =1

P(E’|A₃) = 1 – P(E|A₃)

(iii) that it is of the type A₂ given that a randomly chosen seed does not germinate.

We use Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.

Let events E₁, E₂ be the following:

E₁ be the event that letter is from TATA NAGAR and E₂ be the event that letter is from CALCUTTA

Let E be the event that on the letter, two consecutive letters TA are visible.

Since, the letter has come either from CALCUTTA or TATA NAGAR

When two consecutive letters are visible in the case of TATA NAGAR, we have following set of possible consecutive letters

{TA, AT, TA, AN, NA, AG, GA, AR}

In the case of CALCUTTA, we have following set of possible consecutive letters

{CA, AL, LC, CU, UT, TT, TA}

So, P(E|E₁) is the probability that two consecutive letters are visible when letter came from TATA NAGAR

P(E|E₂) is the probability that two consecutive letters are visible when letter came from CALCUTTA

We have to find the probability that the letter came from TATA NAGAR.

We use Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.

∴

P(E|E₁) is the probability that the letter came from TATA NAGAR

Given: there are 2 bags –

Bag 1: 3 black and 4 white balls

Bag 2: 4 black and 3 white balls

Total balls = 7

Let events E₁, E₂ be the following:

E₁ be the event that bag 1 is selected and E₂ be the event that bag 2 is selected

It is given that a die is thrown.

So, total outcomes = 6

The Law of Total Probability:

In a sample space S, let E₁,E₂,E₃…….E_n be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with E₁,E₂,E₃…….E_n, then

P(A) = P(E₁)P(A/E₁)+ P(E₂)P(A/E₂)+ …… P(E_n)P(A/E_n)

Let “E” be the event that black ball is chosen.

P(E|E₁) is the probability that black ball is chosen from the bag 1.

P(E|E₂) is the probability that black ball is chosen from the bag 2.

So,

So, probability of choosing a black ball is the sum of individual probabilities of choosing the black from the given bags.

From the law of total probability,

P(E) = P(E₁) × P(E|E₁) + P(E₂) × P(E|E₂)

There are 3 urns U₁, U₂ and U₃

U₁ = 2 white and 3 black balls

U₂ = 3 white and 2 black balls

U₃ = 4 white and 1 black balls

Total balls = 5

As there is an equal probability of each urn being chosen

Let E₁, E₂ and E₃ be the event that a ball is chosen from an urn U₁,

U₂ and U₃ respectively.

Now, let A be the event that white ball is drawn.

P(A|E₁) is the probability that white ball is chosen from urn U₁

P(A|E₂) is the probability that white ball is chosen from urn U₂

P(A|E₃) is the probability that white ball is chosen from urn U₃

Now, we have to find the probability that the ball is drawn was from

U₂.

We use Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.

∴

P(E₂|A) is the probability that white ball is selected from urn U₂.

Let events E₁, E₂, E₃ be the following:

E₁ be the event that person has TB and E₂ be the event that the person does not have TB

Total persons = 1000

So,

Now, Let E be the event that the person is diagnosed to have TB

P(E|E₁) is the probability that TB is detected when a person is actually suffering

P(E|E₂) the probability of an healthy person diagnosed to have TB

So, P(E|E₁) = 0.99 and P(E|E₂) = 0.001

Now, we have to find the probability that the person actually has TB

We use Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.

∴

P(E₁|E) is the probability that person actually has TB

[Divide by 9 both numerator and denominator]

Let events E₁, E₂, E₃ be the following:

E₁: event that item is manufactured by machine A

E₂: event that item is manufactured by machine B

E₃: event that item is manufactured by machine C

Clearly, E₁, E₂ and E₃ are mutually exclusive and exhaustive events and hence, they represent a partition of sample space.

Given that:

Items manufactured on machine A = 50%

Items manufactured on machine B = 30%

Items manufactured on machine C = 20%

So,

Now, Let E be the event that ‘an item is defective’.

P(E|E₁) is the probability of the item drawn is defective given that it is manufactured on machine A = 2%

P(E|E₂) is the probability of the item drawn is defective given that it is manufactured on machine B = 2%

P(E|E₃) is the probability of the item drawn is defective given that it is manufactured on machine C = 3%

So,

Now, we have to find the probability that the item which is picked

up is defective, it was manufactured on machine A

We use Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.

∴

P(E₁|E) is the probability that the item is drawn is defective and it was manufactured on machine A

Given:

Thus, we have the probability distribution of X is

(i) the value of k

We know that,

Sum of the probabilities = 1

∴ 2k + 3k + 4k + 5k + 10k + 12k + 14k = 1

⇒ 50k = 1

⇒ k = 0.02

(ii) To find: E(X)

The probability distribution of X is:

Therefore,

μ = E(X)

∴ E(X) = 2k + 6k + 12k + 20k + 50k + 72k + 98k + 0

= 260k

= 5.2 …(i)

(iii) To find: Standard deviation of X

We know that,

Var(X) = E(X²) – [E(X)]²

= ΣX²P(X) – [Σ{XP(X)}]²

= [2k + 12k + 36k + 80k + 250k + 432k + 686k +0] – [5.2]² = 1498k – 27.04

= 29.96 – 27.04

= 2.92

We know that,

standard deviation of X = √Var(X) = √2.92 = 1.7088

≅ 1.7

(i) Given: E(X) = 2.94

We know that, μ = E(X)

[given: E(X) = 2.94]

⇒ 2.94 × 50 = 69 + 26A

⇒ 147 – 69 = 26A

⇒ 78 = 26A

⇒ A = 3

(ii) We know that,

Var(X) = E(X²) – [E(X)]²

= ΣX²P(X) – [Σ{XP(X)}]²

= ΣX²P(X) – (2.94)²

Firstly, we find ΣX²P(X)

=19.06

Now, Var(X) = 19.06 – (2.94)²

= 19.06 – 8.6436

= 10.4164

Given:

We know that,

Sum of the probabilities = 1

⇒ k + 4k + 9k + 8k + 10k + 12k = 1

⇒ 44k = 1

(i) To find: E(X)

We know that, μ = E(X)

or

E(X) = ΣXP(X)

= 1 × k + 2 × 4k + 3 × 9k + 4 × 8k + 5 × 10k + 6 × 12k

= k + 8k + 27k + 32k + 50k + 72k

= 190k

= 4.32

(ii) To find: E(3X²)

Firstly, we find E(X²)

We know that,

E(X²) = ΣX²P(X)

= 1² × k + 2² × 4k + 3² × 9k + 4² × 8k + 5² × 10k + 6² × 12k

= k + 16k + 81k + 128k + 250k + 432k

= 908k

= 20.636

≅ 20.64

∴ E(3X²) = 3 × 20.64 = 61.92

(iii) P(X ≥ 4) = P(X = 4) + P(X = 5) + P(X = 6)

= 8k + 10k + 12k

= 30k

Given: n coins have head on both the sides

and (n + 1) coins are fair coins

Total coins = 2n + 1

Let events E₁, E₂ be the following:

E₁ = Event that an unfair coin is selected

E₂ = Event that a fair coin is selected

The Law of Total Probability:

In a sample space S, let E₁,E₂,E₃…….E_n be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with E₁,E₂,E₃…….E_n, then

P(A) = P(E₁)P(A/E₁)+ P(E₂)P(A/E₂)+ …… P(E_n)P(A/E_n)

Let “E” be the event that the toss result is a head

P(E|E₁) is the probability of getting a head when unfair coin is tossed

P(E|E₂) is the probability of getting a head when fair coin is tossed

So,

From the law of total probability,

∴ P(E) = P(E₁) × P(E|E₁) + P(E₂) × P(E|E₂)

(Given)

⇒ 31 × 2(2n+1) = 42 × (3n + 1)

⇒ 124n + 62 = 126n + 42

⇒ 2n = 20

⇒ n = 10

Hence, the value of n is 10.

Let X denotes a random variable of number of aces

Clearly, X can take values 0, 1 or 2 because only two cards are drawn.

Total deck of cards = 52

and total no. of ACE cards in a deck of cards = 4

Now, since the draws are done without replacement, therefore, the two draws are not independent.

Therefore,

P(X = 0) = Probability of no ace being drawn

= P(non – ace and non – ace)

= P(non – ace) × P(non – ace)

P(X = 1) = Probability that 1 card is an ace

= P(ace and non – ace or non –ace and ace)

= P(ace and non – ace) + P(non – ace and ace) = P(ace) P(non – ace) + P(non – ace) P(ace)

P(X = 2) = Probability that both cards are ace

= P(ace and ace)

= P(ace) × P(ace)

We know that,

Mean (μ) = E(X) = ΣXP(X)

Also, Var(X) = E(X²) – [E(X)]²

= ΣX²P(X) – [E(X)]²

= 0.1629 – 0.0237

= 0.1392

∴Standard Deviation = √Var(X) = √0.1392 ≅ 0.373(approx.)

Let X be the random variable for a ‘success’ for getting an even number on a toss.

∴ X = 0, 1, 2

n = 2

Even number on dice = 2, 4, 6

∴ Total possibility of getting an even number = 3

Total number on dice = 6

p = probability of getting an even number on a toss

q = 1 – p

Now, the probability of x successes in n–Bernoulli trials is ⁿC_rp^rq^n-r

Thus, P( x successes) = ⁿC_rp^rq^n-r

=1

Also, Var(X) = E(X²) – [E(X)]²

= ΣX²P(X) – [E(X)]²

= 1.5 – 1

= 0.5

Here, S = { (1,2),(1,3),(1,4),(1,5)

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

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

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

(5,1),(5,2),(5,3),(5,4)}

Total Sample Space, n(S) = 20

Let random variable be X which denotes the sum of the numbers on the cards drawn.

∴ X = 3, 4, 5, 6, 7, 8, 9

At X = 3

The cards whose sum is 3 are (1,2), (2,1)

At X = 4

The cards whose sum is 4 are (1,3), (3,1)

At X = 5

The cards whose sum is 5 are (1,4),(2,3),(3,2),(4,1)

At X = 6

The cards whose sum is 6 are (1,5), (2,4),(4,2),(5,1)

At X = 7

The cards whose sum is 7 are (2,5),(3,4),(4,3),(5,2)

At X = 8

The cards whose sum is 8 are (3,5), (5,3)

At X = 9

The cards whose sum is 9 are (4,5), (5,4)

∴ Mean, E(X) = ΣXP(X)

= 6

Also,

= 39

Now,

Var X = ΣX²P(X) – [ΣXP(X)]²

= 39 – 36

= 3

We have,

We know that,

P(B|A) × P(A) = P(A∩B)

[Property of Conditional Probability]

Hence, Correct option is C

We have,

We know that,

P(A|B) × P(B) = P(A∩B)

[Property of Conditional Probability]

Hence, Correct option is A

We have,

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

[Additive Law of Probability]

Now, We know that the Property of Conditional Probability:

P(A|B) × P(B) = P(A ∩ B)

…(i)

P(B|A) × P(A) = P(B ∩ A)

…(ii)

Multiplying eq. (i) and (ii), we get

Hence, the Correct option is D

We have,

P(A’|B’) × P(B’) = P(A’ ∩ B’)

…(i)

P(B’|A’) × P(A’) = P(B’ ∩ A’)

…(ii)

On multiplying eq. (i) and (ii), we get

[∵P(A’ ∩ B’) = P[(A ∪ B)’]= 1 – P(A ∪ B)]

[Additive law of Probability]

Hence, the correct option is C

We have,

We know that,

P(A|B) × P(B) = P(A ∩ B)

[Property of Conditional Probability]

Now, P(A’ ∩ B’) = 1 – P(A ∪ B)

= 1 – [P(A) + P(B) – P(A ∩ B)]

[∵P(A ∪ B) = P(A) + P(B) – P(A ∩ B)]

Hence, the correct option is C

We have,

P(A) = 0.4, P(B) = 0.8 and P(B|A) = 0.6

We know that,

P(B|A) × P(A) = P(B ∩ A)

⇒ 0.6 × 0.4 = P(B ∩ A)

⇒ P(B ∩ A) = 0.24

Now,

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

[Additive Law of Probability]

= 0.4 + 0.8 – 0.24

= 0.96

Hence, the correct option is D

Given: A ≠ φ, B ≠φ

CASE 1: If we take option (A) i.e. P(A|B) = P(A).P(B)

LHS =

CASE 2: If we take option (B) i.e.

this is true, we all know that this is conditional probability.

CASE 3: If we take option C .i.e. P(A|B)×P(B|A) = 1

LHS = P(A|B) × P(B|A)

≠ RHS

Hence, the correct option is B

We have,

P(A) = 0.4, P(B) = 0.3 and P(A ∪ B) = 0.5

Now,

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

[Additive Law of Probability]

⇒ 0.5 = 0.4 + 0.3 – P(A ∩ B)

⇒ P(A ∩ B) = 0.7 – 0.5

⇒ P(A ∩ B) = 0.2

∴ P(B’ ∩ A) = P(B’) P(A)

= [1 – P(B)]× P(A)

[sum of the probabilities of an event and its complement is 1]

= P(A) – P(A)P(B)

= P(A) – P(A ∩ B)

= 0.4 – 0.2

= 0.2

Hence, the correct option is D

We have,

Now, We know that

P(A|B) × P(B) = P(A ∩ B)

[Property of conditional Probability]

Now,

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

[Additive Law of Probability]

Hence, the correct option is C

Now, referring the above solution

Hence, the correct option is D

We have,

Now, We know that

P(A|B) × P(B) = P(A ∩ B)

[Property of Conditional Probability]

Now,

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

[Additive Law of Probability]

∴ P(A ∪ B)’ = P[A’ ∩ B’]

= 1 – P(A ∪ B)

and P(A’ ∪ B) = 1 – P(A’ ∩ B)

= 1 – [P(A) – P(A ∩ B)]

= 1

Hence, the correct option is D

Given, P(A) = 7/13, P(B) = 9/13 and P(A ∩ B) = 4/13

From Venn diagram,

Hence,

We have P(A) > 0 and P(B) ≠ 1

By de Morgan’s Law:

P(A'∩B') = P(A∪B)^’

As A and B are independent so A^’ and B^’ are also independent.

P (A’ ∩ B’) =P(A’).P(B’)

And, we know, P(A’) =(1-P(A)) and P(B’)=(1-P(B)

Mutually exclusive are the events which cannot happen at the same time.

For example: when tossing a coin, the result can either be heads or tails but cannot be both.

Events are independent if the occurrence of one event does not influence (and is not influenced by) the occurrence of the other(s).

Eg: Rolling a die and flipping a coin. The probability of getting any number on the die will not affect the probability of getting head or tail in the coin.

So, if A and B are event is independents any information about A can not tell anything about B while if they are mutually exclusive then we know if A occurs B does not occur.

So independent events cannot be mutually exclusive.

Now to test if probability of independent events is 1 or not

Consider an example:

Let A be the event of obtaining a head.

P(A) = 1/2

B be the event of obtaining 5 on a die.

P(B) = 1/6

Now A and B are independent events.

So, P(A) + P(B)

Hence P(A) + P(B)≠ 1

It is true in every case when two events are independent.

Hence option D is correct.

Given, P(A) = 3/8, P(B) = 5/8 and P(A ∪ B) = 3/4

Now, We know that, P(A ∪ B)=P(A)+P(B)- P(A ∩ B)

P(A ∩ B)=

P(A ∩ B)=

So, P(A|B) = , then

P(A|B) =

P(A|B)=2/5

Now, For, P(A’|B)

P(A’|B)=

P(A’|B)=

P(A’|B)= 3/5

Therefore, P(A | B).P(A′|B)=

Hence, P(A | B).P(A′|B)=6/25

If the events A and B are independent, then we know

P(A ∩ B)=P(A).P(B)

Prove: From the definition of the independent Event,

We know,

And, Also

Given, P(E) = 0.3, P(E ∪ F) = 0.5

Also, E and F are independent, then

P (E ∩ F)=P(E).P(F)

We know, P(E ∪ F)=P(E)+P(F)- P(E ∩ F)

P(E ∪ F)=P(E)+P(F)- [P(E) P(F)]

0.5 = 0.3 + P(F)-0.3P(F)

0.5-0.3 =(1- 0.3) P(F)

P(F)=

P(F)=

Since P(E|F)-P(F|E)

P(E|F)-P(F|E)

P(E|F)-P(F|E)=1/70

Probability of getting exactly one red ball

= P(R).P(B).P(B) + P(B).P(R).P(B) + P(B).P(B).P(R)

A bag contains 5 red and 3 blue balls

Total Balls in a Bag = 8

For exactly 1 red ball probability should be

Now, 3 Balls are drawn randomly then possibility for getting 1 red ball

P(E)=P(R).P(B)+P(B).P(R)

P(E) =

Hence, P(E) = 4/7

Here, P(A)=0.4 P(B)=0.3 and P(C)=0.2

And, P(A’)=1-P(A)=[1-0.4] = 0.6

P(B’)=1-P(B)=[1-0.3] = 0.7

P(C’)=1-P(C)=[1-0.2] = 0.8

P(E)=[P(A)×P(B)×P(C’)]+[P(A)×P(B’)×P(C)]+[P(A’)×P(B)×P(C)]

[(0.4×0.3×0.8)+(0.4×0.7×0.2)+(0.6×0.3×0.2)]

[0.96+0.056+0.036]

0.188

Hence, Probability of two hits is 0.188

We can arrange the statement in set as

S={(B,B,B),(G,G,G),(B,G,G),(G,B,G),(G,G,B),(G,B,B),(B,G,B),(B,B,G)}

Let A be Event that a family has at least one girl then,

A={(G,B,B),(B,G,B),(B,B,G),(G,G,B),(B,G,G)(G,B,G),(G,G,G)

Let B be Event that eldest child is girl then,

B={(G,B,B)(G,G,B),(G,B,G),(G,G,G)

Now, (A ∩ B)={(G,B,B),(G,G,B),(G,B,G,)(G,G,G)

Since,

Hence,

Let A be Event for getting number on dice

And, B be Event that a spade card is selected

A={2,4,6}

B={13}

Since, P(A)=

P(B)

We know, If E and F are two independent events then, P(AՌB)=P(A).P(B)

P(A ∩ B)=

Hence, P(E₁∩ E₂)=

There are total 8 balls in box.

P(G) , Probability of green ball

P(B) , Probability of blue ball

_{Now, The probability of drawing 2 green balls and one blue ball is}

_{P(E)=P(G).P(G).P(B)+P(B).P(G).P(G)+P(G).P(B).P(G)}

Hence,

There are total number of batteries , n= 8

Number of dead batteries are = 3

Probability of dead batteries is

Now, If two batteries are selected without replacement and tested

Then, Probability of second battery without replacement is

Required probability =

We know, probability distribution P(X=r)=ⁿC_r (p)^r q^n-r

Here. Total number coin is tossed, n =

The probability of getting head, p =1/2

The probability of getting tail, q = 1/2

Required probability = ⁸C₃

Let A be the event that the sum of numbers on the dice was less than 6

And, B be the event that the sum of numbers on the dice is 3

A={(1,4)(4,1)(2,3)(3,2)(2,2)(1,3)(3,1)(1,2)(2,1)(1,1)

N(A)=10

B={(1,2)(2,1)

n(B)=2

Required probability =

Required probability =

Hence, The probability is

In the binomial distribution, there are 2 outcomes for each trial and there is a fixed number of trials and the probability of success must be the same for all trials.

Number of cards = 52

Number of queen = 4

Probability of queen out of 52 cards =

Now, According to the question,

A deck of card shuffled again with replacement, then

Probability of getting queen is , 4/52

Therefore, The probability , that both cards are queen is ,

Hence, Probability is

In the examination , we have only two option True and False.

So, The probability of getting True is , p= 1/2

And, similarly, probability of getting False is, q=1/2

The total number of Answer in examination , n=10

The probability of guessing correctly at least 8 it means r=8,9,10

We know, probability distribution P(X=r)=ⁿC_r (p)^r q^n-r

P(X=r)=P(r=8)+P(r=9)+P(r=10)

= ¹⁰C₈¹⁰C₉¹⁰C₁₀

[45+10+1]

[

Hence, Probability is

Total number of person , n=5

Total number of swimmers among total person , r=4

Probability of not swimmer , Q=0.3

So, The probability of swimmer , p=1-Q=0.7

We know, probability distribution P(X=r)=ⁿC_r (p)^r q^n-r

P(X) = ⁿC_r

P(X) =

Hence,⁵C₄ (0.7)⁴ (0.3)

Given, Probability distribution table is given,

Now,

We know

K=32

Hence, The value of k is 32

Given, Probability distribution table is given,

Now,

E(X)=[(-4)×(0.1)+(-3)×(0.2)+(-2)×(0.3)+(-1)×(0.2)+(0×0.2)]

E(X)=[-0.4-0.6-0.6-0.2+0]

E(X)=[-1.8]

Hence, E(X) =-1.8

Given, Probability distribution table is given,

Now,

Hence, E(X²) =10

In binomial distribution, we know P(X=r) = ⁿC_r

where q=1-p

Therefore,

= ⁿC_r / ⁿC_r

= /

Since, ⁿC_r= ⁿC_n-r

Accorting to question, this expression is independent of n and r if

Hence, p =1/2

Let A denotes the event that students failed in physics.

According to question: 30% students failed in physics.

∴ P(A) = 0.30

Similarly, if we denote the event of failing in maths with B.

We can write that:

P(B) = 0.25

Also, probability of failing in both subjects can be represented using intersection as shown:

P (A ∩ B) = 0.1

Now we need find a conditional probability of failing of student in physics given that she has failed in mathematics.

We can represent the situation mathematically as-

P(A|B) =?

Using the fundamental idea of conditional probability, we know that:

P(E|F) =

where E & F denotes 2 random events.

∴P(A|B) =

⇒ P(A|B) =

Clearly our answer matches with option B.

∴Option (B) is the only correct choice.

Let E denotes the event that student ‘A’ solves the problem correctly.

∴P(E) = 1/3

Similarly, if we denote the event of ’B’ solving the problem correctly with F

We can write that:

P(F) = 1/4

Note: Observe that both the events are independent.

∴ Probability that both the students solve the question correctly can be represented as-

P (E∩F) == P(E₁) {say} {we can multiply because events are independent}

∴ Probability that both the students could not solve the question correctly can be represented as-

P(E’∩F’) == P(E₂) {say}

Now we are given with some more data and they can be interpreted as:

Given: probability of making a common error and both getting same answer.

Note: If they are making an error, we can be sure that answer coming out is wrong.

Let S denote the event of getting same answer.

∴ above situation can be represented using conditional probability.

P(S|E₂) = 1/20

And if their answer is correct obviously, they will get same answer.

∴P(S|E₁) = 1

We need to find the probability of getting a correct answer if they committed a common error and got the same answer.

Mathematically,

i.e P(E₁|S) = ?

By observing our requirement and availability of equations, we can make guess that Bayes theorem is going to help us.

∴Using Bayes theorem, we get-

P(E₁|S) =

Using the values from above –

P(E₁|S) =

Clearly our answer matches with option D.

∴Option (D) is the only correct choice.

This problem can be solved using Bernoulli trials.

Here n = 5 (as we are drawing 5 pens only)

Success is defined when we get a defective pen.

Let p denotes the probability of success and q probability of failure.

∴ p = 10/100 = 0.1

And q = 1 – 0.1 = 0.9

As we need to find probability of getting at most 1 defective pen.

Let X be a random variable denoting the probability of getting r number of defective pens.

∴ P (drawing atmost 1 defective pen) = P(X = 0) + P(X = 1)

The binomial distribution formula is:

P(x) = _nC_x P^x (1 – P)^{n – x}

Where:

x = total number of “successes.”

P = probability of success on an individual trial

n = number of trials

⇒ P(X = 0) + P(X = 1) =

∴ P(drawing at most 1 defective pen) =

⇒ P(drawing at most 1 defective pen) =

Our answer matches with option D.

∴Option (D) is the only correct choice.

FALSE

For events to be mutually exclusive –

P(A∪B) = P(A) + P(B)

But as per the conditions in question, it is not necessary that they will meet the condition because it might be possible that

P(A ∩ B) ≠ 0

For events to be independent–

P(A ∩ B) = P(A)P(B)

Again P(A) > 0 and P(B)> 0 are not sufficient conditions to validate them.

Tip:you can also check by considering any simple example

TRUE

As A and B are independent

⇒ P(A∩B) = P(A)P(B)

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

As, P(A ∪ B)’ = 1 – P(A ∪ B)

As we know P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

⇒ P(A ∪ B)’ = 1 –[ P(A) + P(B) – P(A ∩ B)]

⇒ P(A’ ∩ B’) = 1 - P(A) - P(B) + P(A)P(B) {as A & B are independent}

= [1 – P(A) ]– P(B) (1-P(A)]

⇒ P(A’∩B’) = (1–P(A))(1–P(B))

= P(A’)P(B’)

hence proved

False

If A and B are mutually exclusive. It implies-

P(A∪B) = P(A) + P(B)

Through this equation we can’t prove in any way that

P(A ∩ B)= P(A)P(B).

So, it is a false statement.

False

If A and B are independent events. It implies-

P(A ∩ B) = P(A)P(B)

Through the above equation we can’t prove in any way that

P(A∪B) = P(A) + P(B)

It is only possible if either P(A) or P(B) = 0,which is not given in question.

So, it is a false statement.

TRUE

If A and B are independent events. It implies-

P(A ∩ B) = P(A)P(B)

Thus, from the definition of independent event we say that statement is true.

TRUE

Mean gives the average of values and if it is related with probability or random variable it is often called expected value.

TRUE

If A and B are independent events. It implies-

P(A ∩ B) = P(A)P(B)

∵ P(A′ ∪ B) = P(A’) + P(B) – P(A’ ∩ B)

and P(A′ ∪ B) represents the probability of event ‘only B’ excluding common points.

From Venn diagram we can see:

∴ P (A′ ∩ B) = P(B) – P (A ∩ B)

⇒ P (A′ ∪ B) = P(A’) + P(B) – P(B) + P (A ∩ B)

⇒ P (A′ ∪ B) = 1 – P(A) + P(A)P(B) {independent events}

⇒ P(A′ ∪ B) = 1 – P(A){1 – P(B)}

⇒ P(A′∪B) = 1 – P(A)P(B’) …proved

TRUE

If A and B are independent events. It implies-

P(A ∩ B) = P(A)P(B)

P(A’ ∩ B) = P(A’)P(B)

And P(A ∩ B’) = P(A)P(B’)

P(exactly one of A, B occurs) = P(A’ ∩ B) + P(A ∩ B’)

⇒ P(exactly one of A, B occurs) = P(A’)P(B)+ P(A)P(B’)

∴Statement is true.

True

As, P(B|A) =

⇒ P(B|A) =

⇒ P(B|A) =

The above equation clearly implies that:

P(B|A) ≥ ;

As we need to add to get the equal term

⇒ LHS is greater.

True

Let A, B,C denotes occurrence of events A,B and C and A’,B’ and C’ not occurrence.

As, P(A) = P(B) = P(C) = p

And P(A’) = P(B’) = P(C’) = p

P (At least two of A, B, C occur) = P(A ∩ B ∩ C’) + P(A ∩ B’ ∩ C) + P(A’ ∩ B ∩ C)

∵ events are independent:

P (At least two of A, B, C occur) = P(A)P(B)P(C’) + P(A)P(B’)P(C)+P(A’)P(B)P(C) = 3p²(1-p) = 3p² – p³

Hence, statement is true.

p = 1/3

As we know that:

P(A|B) =

∴ p =

⇒ 3p = 9p – 2

⇒ 6p = 2

∴ p = 1/3

10/9

P(A’ ∪ B’) = P(A’) + P(B’) – P(A’ ∩ B’) {using union of two sets}

⇒ P(A’) + P(B’) = P(A’ ∪ B’) + P(A’ ∩ B’)

∵ P(A’ ∩ B’) = P(A ∪ B)’ {using De Morgan’s law}

⇒ P(A’ ∩ B’) = 1- P(A ∪ B) = 1 – 5/9 = 4/9

∴P(A’) + P(B’) = 2/3 + 4/9 = 10/9

p = 1/10

As n = 5 {representing no. of trials}

p = probability of success

As it is a binomial distribution.

∴ probability of failure = q = 1 – p

Given,

P(X = 2) = 9.P(X = 3)

The binomial distribution formula is:

P(x) = _nC_x P^x (1 – P)^{n – x}

Where:

x = total number of “successes.”

P = probability of success on an individual trial

n = number of trials

Using binomial distribution,

⇒ ⁵C₂p²q^5-2 = 9⁵C₃p³q^5-3

⇒ 10p²q³ = 9×10p³q²

⇒ 10q = 90p {As , p≠0 and q ≠ 0}

⇒ q = 9p

⇒ 1-p = 9p ⇒ 10p = 1

∴p = 1/10

var(X) = E(X²) – [E(X)]² , where E(X) represents mean or expected value for random variable X

Variance is nothing but the mean of deviation of Random variable from its expected value.

Var(X) = ∑p_i(X_i – X)².

On expanding we get the formula: var(X) = E(X²) – [E(X)]²

Note: You can remember this formula it is a direct formula question.

Independent

As, we know that-

P(A|B) =

but it is given that-

⇒ P(A ∩ B) = P(A)P(B)

This implies that A and B are independent of each other.

∴A is independent of B