Probability

Given: P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2

We know that

By definition of conditional probability,

And

Given: P(B) = 0.5 and P(A ∩ B) = 0.32

We know that

By definition of conditional probability,

Given: P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4

(i) We know that

By definition of conditional probability,

⇒ P(A ∩ B) = P(B|A) P(A)

⇒ P(A ∩ B) = 0.4 × 0.8

⇒ P(A ∩ B) = 0.32

(ii) We know that

By definition of conditional probability,

⇒ P(A|B) = 0.64

(iii) Now, ∵ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

⇒ P(A ∪ B) = 0.8 + 0.5 – 0.32 = 1.3 – 0.32

⇒ P(A ∪ B) = 0.98

Given: and

……….(i)

We know that

By definition of conditional probability,

⇒ P(A ∩ B) = P(A|B) P(B)

……….(ii)

Now, ∵ P(A * B) = P(A) + P(B) – P(A ∩ B)

Given:

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

⇒ P(A ∩ B) = P(A) + P(B) – P(A ∪ B)

(ii) Now, ∵ By definition of conditional probability,

(iii) Again, ∵ By definition of conditional probability,

The sample space of the given experiment will be:

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

(i) Here, E: head on third toss

And F: heads on first two tosses

⇒ E = {HHH, HTH, THH, TTH} and F = {HHH, HHT}

⇒ E ∩ F = {HHH}

So,

Now, we know that

By definition of conditional probability,

(ii) Here, E: at least two heads

And F: at most two heads

⇒ E = {HHH, HHT, HTH, THH} and F = {HHT, HTH, THH, HTT, THT, TTH, TTT}

⇒ E ∩ F = {HHT, HTH, THH}

So,

Now, we know that

By definition of conditional probability,

(iii) Here, E: at most two tails

And F: at least one tail

⇒ E = {HHH, HHT, HTH, THH, HTT, THT, TTH}

And F = {HHT, HTH, THH, HTT, THT, TTH, TTT}

So,

Now, we know that

By definition of conditional probability,

The sample space of the given experiment will be:

S = {HH, HT, TH, TT}

(i) Here, E: tail appears on one coin

And F: one coin shows head

⇒ E = {HT, TH} and F = {HT, TH}

⇒ E ∩ F = {HT, TH}

So,

Now, we know that

By definition of conditional probability,

⇒ P(E|F) = 1

(ii) Here, E: no tail appears

And F: no head appears

⇒ E = {HH} and F = {TT}

⇒ E ∩ F = ϕ

So,

Now, we know that

By definition of conditional probability,

⇒ P(E|F) = 0

The sample space has 216 outcomes, where each element of the sample space has 3 entries and is of the form (x, y, z) where 1 ≤ x, y, z ≤ 6.

Here, E: 4 appears on the third toss

Now, F: 6 and 5 appears respectively on first two tosses

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

⇒ E ∩ F = {(6, 5, 4)}

So,

Now, we know that

By definition of conditional probability,

Let M denote mother, F denote father and S denote son.

Then, the sample space for the given experiment will be:

S = {MFS, SFM, FSM, MSF, SMF, FMS}

Here, E: Son on one end

And F: Father in middle

⇒ E = {MFS, SFM, SMF, FMS} and F = {MFS, SFM}

⇒ E ∩ F = {MFS, SFM}

So,

Now, we know that

By definition of conditional probability,

⇒ P(E|F) = 1

Let B denote black colored die and R denote red colored die.

Then, the sample space for the given experiment will be:

(a) Let A be the event of ‘obtaining a sum greater than 9’ and B be the event of ‘getting a 5 on black die’.

Then, A = {(B4, R6), (B5, R5), (B5, R6), (B6, R4), (B6, R5), (B6, R6)}

And B = {(B5, R1), (B5, R2), (B5, R3), (B5, R4), (B5, R5), (B5, R6)}

⇒ A ∩ B = {(B5, R5), (B5, R6)}

So,

Now, we know that

By definition of conditional probability,

(b) Let A be the event of ‘obtaining a sum 8’ and B be the event of ‘getting a number less than 4 on red die’.

Then, A = {(B2, R6), (B3, R5), (B4, R4), (B5, R3), (B6, R2)}

And

⇒ A ∩ B = {(B5, R3), (B6, R2)}

So,

Now, we know that

By definition of conditional probability,

The sample space for the given experiment will be:

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

Here, E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5} ……….(I)

……….(II)

Now, E ∩ F = {3}, F ∩ G = {2, 3}, E ∩ G = {3, 5} ……….(III)

……….(IV)

(i) We know that

By definition of conditional probability,

[Using (II) and (IV)]

Similarly, we have

[Using (II) and (IV)]

(ii) We know that

By definition of conditional probability,

Similarly, we have

(iii) Clearly, from (I), we have

E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5}

⇒ E ∪ F = {1, 2, 3, 5}

⇒ (E ∪ F) ∩ G = {2, 3, 5}

……….(V)

Now, we know that

By definition of conditional probability,

[Using (II) and (V)]

Similarly, we have E ∩ F = {3} [Using (III)]

And G = {2, 3, 4, 5} [Using (I)]

⇒ (E ∩ F) ∩ G = {3}

……….(VI)

So,

[Using (II) and (VI)]

Let B denote boy and G denote girl.

Then, the sample space of the given experiment will be:

S = {GG, GB, BG, BB}

Let E be the event that ‘both are girls’.

⇒ E = {GG}

(i) Let F be the event that ‘the youngest is a girl’.

⇒ F = {GG, BG}

……….(I)

Now, E ∩ F = {GG}

……….(II)

Now, we know that

By definition of conditional probability,

[Using (I) and (II)]

(ii) Let H be the event that ‘at least one is a girl’.

⇒ H = {GG, GB, BG}

……….(III)

Now, E ∩ H = {GG}

……….(IV)

Now, we know that

By definition of conditional probability,

[Using (III) and (IV)]

Here, there are two types of questions, True/False or Multiple Choice Questions (T/F or MCQ), and each of them are divided into Easy and Difficult type, as shown below in the tree diagram.

So, in all, there are, 500 T/F questions and 900 MCQs.

Also, there are 800 Easy questions and 600 Difficult questions.

⇒ The sample space of this experiment has 500 + 900 = 1400 outcomes.

Now,

Let E be the event of ‘getting an Easy question’ and F be the event of ‘getting an MCQ’.

and ……….(i)

Now, E ∩ F is the event of getting an MCQ which is Easy.

Clearly, from the diagram, we know that there are 500 MCQs that are easy.

So, ……….(ii)

Now, we know that

By definition of conditional probability,

[Using (i) and (ii)]

The sample space of the given experiment is:

Let E be the event that ‘the sum of numbers on the dice is 4’ and F be the event that ‘the two numbers appearing on throwing the two dice are different’.

⇒ E = {(1, 3), (2, 2), (3, 1)}

And

⇒ E ∩ F = {(1, 3), (3, 1)}

……….(i)

Now, we know that

By definition of conditional probability,

The experiment is explained below in the tree diagram:

The sample space of the given experiment is:

Let E be the event that ‘the coin shows a tail’ and F be the event that ‘at least one die shows a 3’.

⇒ E = {1T, 2T, 4T, 5T} and F = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 3)}

⇒ E ∩ F = ϕ ⇒ P(E ∩ F) = 0 ……….(i)

Now, we know that

By definition of conditional probability,

[Using (i)]

⇒ P(E|F) = 0

We know that

By definition of conditional probability,

……….(i)

Given:

And P(B) = 0

⇒ Using (i), we have

, which is not defined.

Given: P(A|B) = P(B|A) ……….(i)

Now, we know that

By definition of conditional probability,

……….(ii)

……….(iii)

Using (i), we have

P(A|B) = P(B|A)

⇒ P(A) = P(B)

Given:

As A and B are independent events.

⇒ P (A ∩ B) = P(A).P(B)

Given: A pack of 52 cards.

As we know there are 26 cards in total which are black. Let A and B denotes respectively the events that the first and second drawn cards are black.

Now, P(A) = P(black card in first draw) =

Because the second card is drawn without replacement so, now the total number of black card will be 25 and total cards will be 51. i.e. the conditional probability of B given that A has already occurred.

Now, P = P(black card in second draw) =

Thus the probability that both the cards are black:

⇒ P(A ∩ B) =

Hence, the probability that both the cards are black = .

Given: A box of oranges.

Let A, B and C denotes respectively the events that the first, second and third drawn orange is good.

Now, P(A) = P(good orange in first draw) =

Because the second orange is drawn without replacement so, now the total number of good oranges will be 11 and total oranges will be 14. i.e. the conditional probability of B given that A has already occurred.

Now, P = P(good orange in second draw) =

Because the third orange is drawn without replacement so, now the total number of good oranges will be 10 and total orangs will be 13. i.e. the conditional probability of C given that A nd B has already occurred.

Now, P= P(good orange in third draw) =

Thus the probability that all the oranges are good:

⇒ P(A ∩ B ∩ C) =

Hence, the probability that a box will be approved for sale

Given: A fair coin and an unbiased die are tossed.

We know that the sample space S:

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)}

Let A be the event ‘head appears on the coin:

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

⇒ P(A) =

Now, Let B be the event 3 on the die:

⇒ B = {(H,3), (T,3)}

As, A ∩ B = {(H,3)}

⇒ P(A ∩ B) = ..(1)

And P(A) . P(B) = ..(2)

From (1) and (2) P (A ∩ B) = P(A) . P(B)

Therefore, A and B are independent events.

Given: A die is given.

The sample space for the dice will be:

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

Let A be the event, the number is even:

⇒ A = {2, 4, 6}

Now, Let B be the event, the number is red:

⇒ B = {1, 2, 3}

⇒ P(B) =

As, A ∩ B = {2}

⇒ P(A ∩ B) = ..(1)

And P(A) . P(B) = ..(2)

From (1) and (2) P (A ∩ B) ≠ P(A) . P(B)

Therefore, A and B are not independent events.

Given: P(E) = , P(F) = and P(E ∩ F) =

P(E) . P(F) =

⇒ P (E ∩ F) ≠ P(E) . P(F)

Therefore, E and F are not independent events.

Given: P(A) =, P (A ∪ B) =and P(B) = p

(i) mutually exclusive

When A and B are mutually exclusive.

Then (A ∩ B) = ϕ

⇒ P (A ∩ B) = 0

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

⇒

⇒

(ii) independent

When A and B are independent.

⇒ P (A ∩ B) = P(A) . P(B)

⇒ P (A ∩ B) =

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

⇒

⇒

⇒

Given: P(A) = 0.3 and P(B) = 0.4

(i) P(A ∩ B)

When A and B are independent.

⇒ P (A ∩ B) = P(A) . P(B)

⇒ P (A ∩ B) = 0.3 × 0.4

⇒ P (A ∩ B) = 0.12

(ii) P(A ∪ B)

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

⇒ P (A ∪ B) = 0.3 + 0.4 – 0.12

⇒ P (A ∪ B) = 0.58

(iii) P (A|B)

As we know

⇒

⇒ P (A|B) = 0.3

(iv) P (B|A)

As we know

⇒

⇒ P (B|A) = 0.4

Given: P (A) =, P (B) = and P (A ∩ B) =

P(not A and not B) = P(A^’ ∩ B^’)

As, { A^’ ∩ B^’ =(A ∪ B)^’}

⇒ P(not A and not B) = P ((A ∪ B)^’)

= 1 - P (A ∪ B)

= 1- [P(A) + P(B) - P (A ∩ B)]

Given: P (A) =, P(B) =and P(not A or not B) = 1/4

⇒ P (A^’∪ B^’) = 1/4

⇒ P (A ∩ B)^’ = 1/4

⇒ 1 - P (A ∩ B) = 1/4

⇒ P (A ∩ B) = 1 - 1/4

⇒ P (A ∩ B) = 3/4 ..(1)

And P(A) . P(B) = ..(2)

From (1) and (2) P (A ∩ B) ≠ P(A) . P(B)

Therefore, A and B are not independent events.

Given: P(A) = 0.3, P(B) = 0.6.

(i) P(A and B)

As A and B are independent events.

⇒ P(A and B) = P (A ∩ B) = P(A) . P(B)

= 0.3 × 0.6

= 0.18

(ii) P(A and not B)

⇒ P(A and not B) = P (A ∩ B^’) = P(A) - P(A ∩ B)

= 0.3 - 0.18

= 0.12

(iii) P(A or B)

⇒ P(A or B) = P(A ∪ B)

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

⇒ P (A ∪ B) = 0.3 + 0.6 – 0.18

⇒ P (A ∪ B) = 0.72

(iv) P(neither A nor B)

P(neither A nor B) = P(A^’ ∩ B^’)

As, { A^’ ∩ B^’ =(A ∪ B)^’}

⇒ P(neither A nor B) = P ((A ∪ B)^’)

= 1 - P (A ∪ B)

= 1 - 0.72

= 0.28

Given: A die is tossed thrice.

The sample space S = {1, 2, 3, 4, 5, 6}

Let P(A) = probability of getting an odd number in first throw.

⇒ P(A) = .

Let P(B) = probability of getting an even number.

⇒ P(B) = .

Now, probability of getting an even number in three times =

So, probability of getting an odd number at least once

= 1 – probability of getting an odd number in no throw

= 1 - probability of getting an even number in three times

∴ probability of getting an odd number at least once =

Given: A box containing 10 black and 8 red balls.

Total number of balls in box = 18

(i) both balls are red.

Probability of getting a red ball in first draw =

As the ball is replaced after first throw,

Hence, Probability of getting a red ball in second draw =

Now, Probability of getting both balls red =

(ii) first ball is black and second is red.

Probability of getting a black ball in first draw =

As the ball is replaced after first throw,

Hence, Probability of getting a red ball in second draw =

Now, Probability of getting first ball is black and second is red =

(iii) one of them is black and other is red.

Probability of getting a black ball in first draw =

As the ball is replaced after first throw,

Hence, Probability of getting a red ball in second draw =

Now, Probability of getting first ball is black and second is red =

Probability of getting a red ball in first draw =

As the ball is replaced after first throw,

Hence, Probability of getting a black ball in second draw =

Now, Probability of getting first ball is red and second is black =

Therefore, Probability of getting one of them is black and other is red :

= Probability of getting first ball is black and second is red + Probability of getting first ball is red and second is black

=

Given:

P(A) = Probability of solving the problem by A = 1/2

P(B) = Probability of solving the problem by B = 1/3

Because A and B both are independent.

⇒ P (A ∩ B) = P(A) . P(B)

⇒ P (A ∩ B) =

P(A^’) = 1 – P(A) = 1 – 1/2 = 1/2

P(B^’) = 1 – P(B) =

(i) the problem is solved

The problem is solved, i.e. it is either solved by A or it is solved by B.

= P(A ∪ B)

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

⇒ P (A ∪ B) =

(ii) exactly one of them solves the problem

i.e. either problem is solved by A but not by B or vice versa

i.e. P(A).P(B^’) + P(A^’).P(B)

=

=

⇒ P(A).P(B^’) + P(A^’).P(B) = 1/2

Given: A deck of 52 cards.

(i) In a deck of 52 cards, 13 cards are spade and 4 cards are ace and only one card is there which is spade and ace both.

Hence, P(E) = The card drawn is a spade =

P(F) = The card drawn is an ace =

P(E ∩ F) = The card drawn is a spade and ace both = ..(1)

And P(E) . P(F)

..(2)

_{From (1) and (2)}

⇒ P (E ∩ F) = P(E) . P(F)

Hence, E and F are independent events.

(ii) In a deck of 52 cards, 26 cards are black and 4 cards are king and only two card are there which are black and king both.

Hence, P(E) = The card drawn is of black =

P(F) = The card drawn is a king =

P(E ∩ F) = The card drawn is a black and king both = ..(1)

And P(E) . P(F)

..(2)

_{From (1) and (2)}

⇒ P (E ∩ F) = P(E) . P(F)

Hence, E and F are independent events.

(iii) In a deck of 52 cards, 4 cards are queen, 4 cards are king and 4 cards are jack.

Hence, P(E) = The card drawn is either king or queen =

P(F) = The card drawn is either queen or jack =

There are 4 cards which are either king or queen and either queen or jack.

P(E ∩ F) = The card drawn is either king or queen and either queen or jack = ..(1)

And P(E) . P(F)

..(2)

_{From (1) and (2)}

⇒ P (E ∩ F) ≠ P(E) . P(F)

Hence, E and F are not independent events.

Given:

Let H and E denote the number of students who read Hindi and English newspaper respectively.

Hence, P(H) = Probability of students who read Hindi newspaper=

P(E) = Probability of students who read English newspaper =

P (H ∩ E) = Probability of students who read Hindi and English both newspaper =

(a) Find the probability that she reads neither Hindi nor English newspapers.

P(neither H nor E)

P(neither H nor E) = P(H^’ ∩ E^’)

As, { H^’ ∩ E^’ =(H ∪ E)^’}

⇒ P(neither A nor B) = P ((H ∪ E)^’)

= 1 - P (H ∪ E)

= 1- [P(H) + P(E) - P (H ∩ E)]

(b) If she reads Hindi newspaper, find the probability that she reads English newspaper.

P (E|H) = hindi newspaper reading has already occurred and the probability that she reads English newspaper is to find.

As we know

⇒

⇒ P (E|H) =

(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.

P (H|E) = English newspaper reading has already occurred and the probability that she reads Hindi newspaper is to find.

As we know

⇒

⇒ P (H|E) =

Given: A pair of dice is rolled.

Hence the number of outcomes = 36

Let P(E) be the probability to get an even prime number on each die.

As we know the only even prime number is 2.

So, E = {2,2}

⇒

The correct answer is D.

Given: Two events A and B will be independent.

As A and B are independent events.

⇒ P (A ∩ B) = P(A) . P(B)

We solve it using options.

Let P(A) = a, P(B) = b

As, A and B are mutually exclusive

P (A ∩ B) = ϕ

Now, P(A).P(B) = a.b ≠ P(A ∩ B)

⇒ P (A ∩ B) ≠ P(A) . P(B)

Hence, it shows A and B are not Independent events.

(B) P(A′B′) = [1 – P(A)] [1 – P(B)]

⇒ P(A′∩ B′) = 1 – P(A) – P(B) + P(A)P(B)

⇒ 1 - P (A ∪ B) =1 – P(A) – P(B) + P(A)P(B)

= - [P(A) + P(B) - P (A ∩ B)] = – P(A) – P(B) + P(A)P(B)

= - P(A) - P(B) + P (A ∩ B) = – P(A) – P(B) + P(A)P(B)

⇒ P (A ∩ B) = P(A) . P(B)

Hence, it shows A and B are Independent events.

(C) P(A) = P(B)

As, P(A) = P(B)

Let we take the example of a coin

P(A) = probability of getting head = 1/2

P(B) = probability of getting tail = 1/2

A ∩ B = ϕ

P(A ∩ B) = probability of getting head and tail both = 0

Now, P(A).P(B) = 1/2 . 1/2 = 1/4 ≠ P(A ∩ B)

⇒ P (A ∩ B) ≠ P(A) . P(B)

Hence, it shows A and B are not Independent events.

(D) P(A) + P(B) = 1

Let we take the example of a coin

P(A) = probability of getting head = 1/2

P(B) = probability of getting tail = 1/2

Now, P(A) + P(B) = 1/2 + 1/2 = 1

But it doesnot inferred that A and B are independent.

Hence the correct option is B.

Given: An urn contains 5 red and 5 black balls.

Let in first attempt the ball drawn is of red colour.

⇒ P (probability of drawing a red ball)

Now the two balls of same colour (red) are added to the urn then the urn contains 7 red and 5 black balls.

⇒ P (probability of drawing a red ball)

Now Let in first attempt the ball drawn is of black colour.

⇒ P (probability of drawing a black ball)

Now the two balls of same colour (black) are added to the urn then the urn contains 5 red and 7 black balls.

⇒ P (probability of drawing a red ball)

Therefore, the probability of drawing the second ball as of red colour is:

Given: let E₁ be the event of choosing the bag I, E₂ be the event of choosing the another bag say bag II and A be the event of drawing a red ball.

Then P (E₁) = P (E₂) = 1/2

Also P(A|E₁) = P (drawing a red ball from bag I)

And P(A|E₂) = P (drawing a red ball from bag II)

Now the probability of drawing a ball from bag I, being given that it is red, is P(E₁|A).

By using bayes’ theorem, we have:

Given: let E₁ be the event that student is a hostler, E₂ be the event that student is a day scholar and A be the event of getting A grade.

Then

and

Also P(A|E₁) = P (students who attain A grade reside in hostel) = 30% = 0.3

And P(A|E₂) = P (students who attain A grade is day scholar) = 20% = 0.2

Now the probability of students who reside in hostel, being given he attain A grade, is P(E₁|A).

By using bayes’ theorem, we have:

Given: let E₁ be the event that the student knows the answer, E₂ be the event that the student guess the answer and A be the event that the answer is correct.

Then

And

Also P(A|E₁) = P (correct answer given that he knows) = 1

And P(A|E₂) = P (correct answer given that he guesses)

Now the probability that he knows the answer, being given that answer is correct, is P(E₁|A).

By using bayes’ theorem, we have:

Given: let E₁ be the event that person has a disease, E₂ be the event that person don not have a disease and A be the event that blood test is positive.

As E₁ and E₂ are the events which are complimentary to each other.

Then P (E₁) + P (E₂) = 1

⇒ P (E₂) = 1 - P (E₁)

Then

and P (E₂) = 1 – 0.001 = 0.999

Also P(A|E₁) = P (result is positive given that person has disease) = 99% = 0.99

And P(A|E₂) = P (result is positive given that person has no disease) = 0.5% = 0.005

Now the probability that person has a disease, give that his test result is positive is P(E₁|A).

By using bayes’ theorem, we have:

Given: let E₁ be the event of choosing a two headed coin, E₂ be the event of choosing a biased coin and E₃ be the event of choosing an unbiased coin. Let A be the event that the coin shows head.

Then P (E₁) = P (E₂) = P (E₃) = 1/3

As we a headed coin has head on both sides so it will shows head.

Also P(A|E₁) = P (correct answer given that he knows) = 1

And P(A|E₂) = P (coin shows head given that the coin is biased)

And P(A|E₃) = P (coin shows head given that the coin is unbiased)

Now the probability that the coin is two headed, being given that it shows head, is P(E₁|A).

By using bayes’ theorem, we have:

Given: let E₁ be the event that the driver is a scooter driver, E₂ be the event that the driver is a car driver and E₃ be the event that the driver is a truck driver. Let A be the event that the person meet with an accident.

Total number of drivers = 2000 + 4000 + 6000 = 12000

Then

And

And

As we a headed coin has head on both sides so it will shows head.

Also P(A|E₁) = P (accident of a scooter driver)

And P(A|E₂) = P (accident of a car driver)

And P(A|E₃) = P (accident of a truck driver)

Now the probability that the driver is a scooter driver, being given that he met with an accident, is P(E₁|A).

By using bayes’ theorem, we have:

Given: let E₁ be the event that item is produced by A, E₂ be the event that item is produced by B and X be the event that produced product is found to be defective.

Then

and

Also P(X|E₁) = P (item is defective given that it is produced by machine A)

And P(X|E₂) = P (item is defective given that it is produced by machine B)

Now the probability that item is produced by B, being given that item is defective, is P(E₂|A).

By using bayes’ theorem, we have:

Given: let E₁ be the event that first group wins the competition, E₂ be the event that that second group wins the competition and A be the event of introducing a new product.

Then P(E₁) = 0.6

and P(E₂) = 0.4

Also P(A|E₁) = P (introducing a new product given that first group wins) = 0.7

And P(A|E₂) = P (introducing a new product given that second group wins) = 0.3

Now the probability of that new product introduced was by the second group, being given that a new product was introduced, is P(E₂|A).

By using bayes’ theorem, we have:

Given: : let E₁ be the event that the outcome on the die is 5 or 6, E₂ be the event that the outcome on the die is 1, 2, 3 or 4 and A be the event getting exactly head.

Then

And

As in throwing a coin three times we get 8 possibilities.

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

⇒ P(A|E₁) = P (obtaining exactly one head by tossing the coin three times if she get 5 or 6)

And P(A|E₂) = P (obtaining exactly one head by tossing the coin three times if she get 1,2 ,3 or 4)

Now the probability that the girl threw 1, 2, 3 or 4 with a die, being given that she obtained exactly one head, is P(E₂|A).

By using bayes’ theorem, we have:

Given: let E₁ be the event of time consumed by machine A, E₂ be the event of time consumed by machine B and E₃ be the event of time consumed by machine C. Let X be the event of producing defective items.

Then

And

And

As we a headed coin has head on both sides so it will shows head.

Also P(X|E₁) = P (defective item produced by A)

And P(X|E₂) = P (defective item produced by B)

And P(X|E₃) = P (defective item produced by C)

Now the probability that item produced by machine A, being given that defective item is produced, is P(E₁|A).

By using bayes’ theorem, we have:

Given: let E₁ be the event that the drawn card is a diamond, E₂ be the event that the drawn card is not a diamond and A be the event that the card is lost.

As we know, out of 52 cards, 13 cards are diamond and 39 cards are not diamond.

Then

And

Now, when a diamond card is lost then there are 12 diamond cards out of total 51 cards.

Two diamond cards can be drawn out of 12 diamond cards in ¹²C₂ ways.

Similarly, Two diamond cards can be drawn out of total 51 cards in ⁵¹C₂ ways.

Then probability of getting two cards, when one diamond card is lost, is P(A|E₁).

Also P(A|E₁) =¹²C₂ / ⁵¹C₂

Now, when not a diamond card is lost then there are 13 diamond cards out of total 51 cards.

Two diamond cards can be drawn out of 13 diamond cards in ¹³C₂ ways.

Similarly, Two diamond cards can be drawn out of total 51 cards in ⁵¹C₂ ways.

Then probability of getting two cards, when card is lost which is not diamond, is P(A|E₂).

Also P(A|E₂) =¹³C₂ / ⁵¹C₂

Now the probability that the lost card is diamond, being given that the card is lost, is P(E₁|A).

By using Bayes’ theorem, we have:

Given: let E₁ be the event that A speaks truth, E₂ be the event that A lies and X be the event that it appears head.

Therefore,

As E₁ and E₂ are the events which are complimentary to each other.

Then P (E₁) + P (E₂) = 1

⇒ P (E₂) = 1 - P (E₁)

⇒ P (E₂)

If a coin is tossed it may show head or tail.

Hence the probability of getting head is 1/2 whether A speaks a truth or A lies.

P(X|E₁) = P(X|E₂) = 1/2

Now the probability that actually there was head, give that A speaks a truth is P(E₁|X).

By using bayes’ theorem, we have:

Therefore correct answer is (A).

Given: A and B are two events such that A ⊂ B and P(B) ≠ 0

As A ⊂ B ⇒ A ∩ B = A

⇒ P(A ∩ B) = P(A)

As A ⊂ B ⇒ P(A) < P(B)

As we know

consider

it is also known that P(B) ≤ 1

⇒ P(A|B) ≥ P(A) …(3)

Hence, the correct answer is (C).

Given: A given table with values for X and P(X).

As we know the sum of all the probabilities in a probability distribution of a random variable must be one.

Hence the sum of probabilities of given table = 0.4 + 0.4 + 0.2

= 1

Hence, the given table is the probability distributions of a random variable.

Given: A given table with values for X and P(X).

As we see from the table that P(X) = -0.1 for X = 3.

It is known that probability of any observation must always be positive that it can’t be negative.

Hence, the given table is not the probability distributions of a random variable.

Given: A given table with values for X and P(X).

As we know the sum of all the probabilities in a probability distribution of a random variable must be one.

Hence the sum of probabilities of given table = 0.6 + 0.1 + 0.2

= 0.9 ≠ 1

Hence, the given table is not the probability distributions of a random variable.

Given: A given table with values for X and P(X).

As we know the sum of all the probabilities in a probability distribution of a random variable must be one.

Hence the sum of probabilities of given table = 0.3 + 0.2 + 0.4 + 0.1 + 0.05

= 1.05 ≠ 1

Hence, the given table is not the probability distributions of a random variable.

Given: An urn containing 5 red and 2 black balls.

Let R represent red ball and B represent black ball.

Two balls are drawn randomly. Hence, the sample space of the experiment is S = {BB, BR, RB, RR}

X represents the number of black balls.

⇒ X(BB) = 2

X(BR) = 1

X(RB) = 1

X(RR) = 0

Therefore, X is a function on sample space whose range is {0, 1, 2}.

Thus, X is a random variable which can take the values 0, 1 or 2.

Given: A coin is tossed 6 times.

X represents the difference between the number of heads and the number of tails.

⇒ X(6H, 0T) = |6-0| = 6

X(5H, 1T) = |5-1| = 4

X(4H, 2T) = |4-2| = 2

X(3H, 3T) = |3-3| = 0

X(2H, 4T) = |2-4| = 2

X(1H, 5T) = |1-5| = 4

X(0H, 6T) = |0-6| = 6

Therefore, X is a function on sample space whose range is {0, 2, 4, 6}.

Thus, X is a random variable which can take the values 0, 2, 4 or 6.

Given:

A coin is tossed twice. Hence, the sample space of the experiment is S = {HH, HT, TH, TT}

X represents the number of heads.

⇒ X(HH) = 2

X(HT) = 1

X(TH) = 1

X(TT) = 0

Therefore, X is a function on sample space whose range is {0, 1, 2}.

Thus, X is a random variable which can take the values 0, 1 or 2.

As we know,

P(HH) = P(HT) = P(TH) = P(TT) = 1/4

P(X = 0) = P(TT) = 1/4

P(X = 1) = P(HT) + P(TH) = 1/4 + 1/4 = 1/2

P(X = 2) = P(HH) = 1/4

Hence, the required probability distribution is,

three coins are tossed simultaneously. Hence, the sample space of the experiment is S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}

X represents the number of tails.

As we see, X is a function on sample space whose range is {0, 1, 2, 3}.

Thus, X is a random variable which can take the values 0, 1, 2 or 3.

P(X = 0) = P(HHH)

P(X = 1) = P(HHT) + P(HTH) + P(THH)

P(X = 2) = P(HTT) + P(THT) + P(TTH)

P(X = 3) = P(TTT)

Hence, the required probability distribution is,

Four tosses of a coin. Hence, the sample space of the experiment is S = {HHHH, HHHT, HHTH, HTHH, HTTH, HTHT, HHTT, HTTT, THHH, TTHH, THTH, THHT, THTT, TTHT, TTTH, TTTT}

X represents the number of heads.

As we see, X is a function on sample space whose range is {0, 1, 2, 3, 4}.

Thus, X is a random variable which can take the values 0, 1, 2, 3 or 4.

P(X = 0) = P(TTTT)

P(X = 1) = P(HTTT) + P(TTTH) + P(THTT) + P(TTHT)

P(X = 2) = P(HHTT) + P(THHT) + P(TTHH) + P(THTH) + P(HTHT) + P(HTTH)

P(X = 3) = P(THHH) + P(HHHT) + P(HTHH) + P(HHTH)

P(X = 4) = P(HHHH)

Hence, the required probability distribution is,

Given: A die is tossed two times.

When a die is tossed two times then the number of observations will be (6 × 6) = 36.

Now, let X is a random variable which represents the success.

(i) Here success is given as the number greater than 4.

Now

P(X = 0) = P(number ≤ 4 in both tosses)

P(X = 1) = P(number ≤ 4 in first toss and number ≥ 4 in second case) + P(number ≥ 4 in first toss and number ≤ 4 in second case) is:

P(X = 2) = P(number ≥ 4 in both tosses)

Hence, the required probability distribution is,

(ii) Here success is given as six appears on at least one die.

Now

P(X = 0) = P(six does not appear on any of die)

P(X = 1) = P(six appears atleast once of the die)

Hence, the required probability distribution is,

Given: A lot of 30 bulbs which include 6 defectives.

Then number of non-defective bulbs = 30 – 6 = 24

As 4 bulbs are drawn at random with replacement.

Let X denotes the number of defective bulbs from the selected bulbs.

Clearly, X can take the value of 0, 1, 2, 3 or 4.

P(X = 0) = P(4 are non defective and 0 defective)

P(X = 1) = P(3 are non defective and 1 defective)

P(X = 2) = P(2 are non defective and 2 defective)

P(X = 3) = P(1 are non defective and 3 defective)

P(X = 4) = P(0 are non defective and 4 defective)

Hence, the required probability distribution is,

Given: head is 3 times as likely to occur as tail.

Now, let the probability of getting a tail in the biased coin be x.

⇒ P(T) = x

And P(H) = 3x

For a biased coin, P(T) + P(H) = 1

⇒ x + 3x = 1

⇒ 4x = 1

⇒ x = 1/4

Hence, P(T) = 1/4 and P(H) = 3/4

As the coin is tossed twice, so the sample space is {HH, HT, TH, TT}

Let X be a random variable representing the number of tails.

Clearly, X can take the value of 0, 1 or 2.

P(X = 0) = P(no tail) = P(H) × P(H)

P(X = 1) = P(one tail) = P(HT) × P(TH)

P(X = 2) = P(two tail) = P(T) × P(T)

Hence, the required probability distribution is,

Given: A random variable X with its probability distribution.

(i) As we know the sum of all the probabilities in a probability distribution of a random variable must be one.

Hence the sum of probabilities of given table:

⇒ 0 + k + 2k + 2k + 3k + k² + 2k² + 7K² + k = 1

⇒ 10K² + 9k = 1

⇒ 10K² + 9k – 1 = 0

⇒ (10K-1)(k + 1) = 0

It is known that probability of any observation must always be positive that it can’t be negative.

So

(ii) P(X < 3) = ?

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

= 0 + k + 2k

= 3k

(iii) P(X > 6) = ?

P(X > 6) = P(X = 7)

= 7K² + k

(iv) P(0 < X < 3) = ?

P(0 < X < 3) = P(X = 1) + P(X = 2)

= k + 2k

= 3k

Given: A random variable X with its probability distribution.

(a) As we know the sum of all the probabilities in a probability distribution of a random variable must be one.

Hence the sum of probabilities of given table:

⇒ k + 2k + 3k + 0 = 1

⇒ 6k = 1

(b) (i) P(X < 2) = ?

P(X < 2) = P(X = 0) + P(X = 1)

= k + 2k

= 3k

(ii) P(X ≤ 2) = ?

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

= k + 2k + 3k

= 6k

(iii) P(X ≥ 2) = ?

P(X ≥ 2) = P(X = 2) + P(X > 2)

= 3k + 0

= 3k

Given: A coin is tossed three times.

three coins are tossed simultaneously. Hence, the sample space of the experiment is S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}

X represents the number of heads.

As we see, X is a function on sample space whose range is {0, 1, 2, 3}.

Thus, X is a random variable which can take the values 0, 1, 2 or 3.

P(X = 0) = P(TTT)

P(X = 1) = P(TTH) + P(THT) + P(HTT)

P(X = 2) = P(THH) + P(HTH) + P(HHT)

P(X = 3) = P(HHH)

Hence, the required probability distribution is,

Therefore mean μ is:

Given: A die is thrown two times.

When a die is tossed two times then the number of observations will be (6 × 6) = 36.

Now, let X is a random variable which represents the success and is given as six appears on at least one die.

Now

P(X = 0) = P(six does not appear on any of die)

P(X = 1) = P(six appears atleast once of the die)

P(X = 2) = P(six does appear on both of die)

Hence, the required probability distribution is,

Therefore Expectation of X E(X):

Given: first six positive integers.

Two numbers can be selected at random (without replacement) from the first six positive integer in 6 × 5 = 30 ways.

X denote the larger of the two numbers obtained. Hence, X can take any value of 2, 3, 4, 5 or 6.

For X = 2, the possible observations are (1, 2) and (2, 1)

For X = 3, the possible observations are (1, 3), (3,1), (2,3) and (3, 2).

For X = 4, the possible observations are (1, 4), (4, 1), (2,4), (4,2), (3,4) and (4,3).

For X = 5, the possible observations are (1, 5), (5, 1), (2,5), (5,2), (3,5), (5,3) (5, 4) and (4,5).

For X = 6, the possible observations are (1, 6), (6, 1), (2,6), (6,2), (3,6), (6,3) (6, 4), (4,6), (5,6) and (6,5).

Hence, the required probability distribution is,

Therefore E(X) is:

Given: Two fair dice are rolled

When two fair dice are rolled then number of observations will be 6 × 6 = 36.

X denote the sum of the numbers obtained when two fair dice are rolled. Hence, X can take any value of 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 or 12.

For X = 2, the possible observations are (1, 1).

For X = 3, the possible observations are (1,2) and (2,1)

For X = 4, the possible observations are (1,3), (2,2) and (3,1).

For X = 5, the possible observations are (1, 4), (4, 1), (2,3) and (3,2)

For X = 6, the possible observations are (1, 5), (5, 1), (2,4), (4,2) and (3,3).

For X = 7, the possible observations are (1, 6), (6, 1), (2,5), (5,2),(3,4) and (4,3).

For X = 8, the possible observations are (2,6), (6,2),(3,5), (5,3) and (4,4).

For X = 9, the possible observations are (5, 4), (4, 5), (3,6) and (6,3)

For X = 10, the possible observations are (5,5), (4,6) and (6,4).

For X = 11, the possible observations are (6,5) and (5,6)

For X = 12, the possible observations are (6, 6).

Hence, the required probability distribution is,

Therefore E(X) is:

⇒ E(X) = 7

And E(X²) is:

⇒ E(X²) = 54.833

Then Variance, Var(X) = E(X²) – (E(X))²

= 54.833 – (7)²

= 54.833 – 49 = 5.833

And Standard deviation

⇒ Standard deviation = 2.415

Given: The class of 15 students with their ages.

Form the given information we can draw a table:

P(X = 14)

P(X = 15)

P(X = 16)

P(X = 17)

P(X = 18)

P(X = 19)

P(X = 20)

P(X = 21)

Hence, the required probability distribution is,

Therefore E(X) is:

⇒ E(X) = 17.53

And E(X²) is:

⇒ E(X²) = 312.2

Then Variance, Var(X) = E(X²) – (E(X))²

= 312.2 – (17.53)²

= 312.2 – 307.417 ≈ 4.78

And Standard deviation

⇒ Standard deviation ≈ 2.19

Given: X = 0 if members oppose, and X = 1 if members are in favour.

P(X = 0)

P(X = 1)

Hence, the required probability distribution is,

Therefore E(X) is:

= 0 × 0.3 + 1 × 0.7

⇒ E(X) = 0.7

And E(X²) is:

= (0)² × 0.3 + (1)² × 0.7

⇒ E(X²) = 0.7

Then Variance, Var(X) = E(X²) – (E(X))²

= 0.7 – (0.7)²

= 0.7 – 0.49 = 0.21

Given: A die having written 1 on three faces, 2 on two faces and 5 on one face.

Let X be the random variable representing a number on given die.

Then X can take any value of 1, 2 or 5.

The total numbers is six.

Now

P(X = 1)

P(X = 2)

P(X = 5)

Hence, the required probability distribution is,

Therefore Expectation of X E(X):

⇒ E(X) = 2

Hence, the correct answer is (B).

Given: A deck of cards.

X be the number of aces obtained.

Hence, X can take value of 0, 1 or 2.

As we know, in a deck of 52 cards, 4 cards are aces. Therefore 48 cards are non- ace cards.

P(X = 0) = P(0 ace and 2 non ace cards)

P(X = 1) = P(1 ace and 1 non ace cards)

P(X = 2) = = P(2 ace and 0 non ace cards)

Hence, the required probability distribution is,

Therefore Expectation of X E(X):

Hence, the correct answer is (D).

We know that the repeated tosses of a dice are known as Bernouli trials.

Let the number of successes of getting an odd number in an experiment of 6 trials be x.

Probability of getting an odd number in a single throw of a dice(p)

Thus,

Now, here x has a binomial distribution.

Thus, P(X = x) = ⁿC_xq^n-xp^x , where x = 0, 1, 2, …n

= ⁶C_x (1/2)^6-x(1/2)^x

= ⁶C_x (1/2)⁶

(i) Probability of getting 5 successes = P(X = 5)

= ⁶C₅(1/2)⁶

(ii) Probability of getting at least 5 successes = P(X ≥ 5)

= P(X = 5) + P(X = 6)

= ⁶C₅(1/2)⁶ + ⁶C₅ (1/2)⁶

(iii) Probability of getting at most 5 successes = P(X ≤ 5)

We can also write it as: 1 – P(X>5)

= 1 – P(X = 6)

= 1 – ⁶C₆ (1/2)⁶

We know that the repeated tosses of a pair of dice are known as Bernouli trials.

Let the number of times of getting doublets in an experiment of throwing two dice simultaneously four times be x.

Thus, P(X = x) = ⁿC_xq^n-xp^x , where x = 0, 1, 2, …n

= ⁴C_x (5/6)^4-x(1/6)^x

= ⁴C_x (5^4-x/6⁶)

Hence, Probability of getting 2 successes = P(X = 2)

= ⁴C₂(5^4-2/6⁴)

Let there be x number of defective items in a sample of ten items drawn successively.

Now, as we can see that the drawing of the items is done with replacement. Thus, the trials are Bernoulli trials.

Now, probability of getting a defective item,

∴ we can say that x has a binomial distribution, where

Thus, P(X = x) = ⁿC_xq^n-xp^x , where x = 0, 1, 2, …n

Probability of getting not more than one defective item = P(X ≤1)

= P(X = 0) + P(X = 1)

= ¹⁰C₀ (19/20)¹⁰(1/20)⁰ +¹⁰C₁ (19/20)⁹(1/20)¹

Let the number of spade cards among the five drawn cards be x.

As we can observe that the drawing of cards is with replacement, thus, the trials will be Bernoulli trials.

Now, we know that in a deck of 52 cards there are total 13 spade cards.

Thus, Probability of drawing a spade from a deck of 52 cards

Thus, x has a binomial distribution with

Thus, P(X = x) = ⁿC_xq^n-xp^x , where x = 0, 1, 2, …n

(i) Probability of drawing all five cards as spades = P(X = 5)

(ii) Probability of drawing three out five cards as spades = P(X = 3)

(iii) Probability of drawing all five cards as non-spades = P(X = 0)

Let us assume that the number of bulbs that will fuse after 150 days of use in an experiment of 5 trials be x.

As we can see that the trial is made with replacement, thus, the trials will be Bernoulli trials.

It is already mentioned in the question that, p = 0.05

Thus, q = 1 – p = 1 – 0.05 = 0.95

Here, we can clearly observe that x has a binomial representation with n = 5 and p = 0.05

Thus, P(X = x) = ⁿC_xq^n-xp^x , where x = 0, 1, 2, …n

= ⁵C_x(0.95)^5-x(0.05)^x

(i) Probability of no such bulb in a random drawing of 5 bulbs = P(X = 0)

= ⁵C₀(0.95)^5-0(0.05)⁰

= 1× 0.95⁵

= (0.95)⁵

(ii) Probability of not more than one such bulb in a random drawing of 5 bulbs = P(X≤ 1)

= P(X = 0) + P(X = 1)

= ⁵C₀(0.95)^5-0(0.05)⁰+ ⁵C₁(0.95)^5-1(0.05)¹

= 1× 0.95⁵ + 5 × (0.95)⁴ × 0.05

= (0.95)⁴ (0.95 +0.25)

= (0.95)⁴ × 1.2

(iii) Probability of more than one such bulb in a random drawing of 5 bulbs = P(X>1)

= 1 – P(X ≤ 1)

= 1 – [(0.95)⁴ × 1.2]

(iv) Probability of at least one such bulb in a random drawing of 5 bulbs = P(X ≥ 1)

= 1 – P(X < 1)

= 1 – P(X = 0)

= 1 –(0.95)⁵

Let us assume that number of balls with digit marked as zero among the experiment of 4 balls drawn simultaneously be x.

As we can see that the balls are drawn with replacement, thus, the trial is a Bernoulli trial.

Probability of a ball drawn from the bag to be marked as digit

It can be clearly observed that X has a binomial distribution with

Thus, P(X = x) = ⁿC_xq^n-xp^x , where x = 0, 1, 2, …n

Probability of no ball marked with zero among the 4 balls = P(X = 0)

Let us assume that the number of correctly answered questions out of twenty questions be x.

Since, ‘head’ on the coin shows the true answer and the ‘tail’ on the coin shows the false answers. Thus, the repeated tosses or the correctly answered questions are Bernoulli trails.

Here, it can be clearly observed that x has binomal distribution, where

Thus, P(X = x) = ⁿC_xq^n-xp^x , where x = 0, 1, 2, …n

Probability of at least 12 questions answered correctly = p(X ≥ 12)

= P(X = 12) +P(X = 13)+…+P(X = 20)

As per the question,

X is any random variable whose binomial distribution is

Thus, P(X = x) = ⁿC_xq^n-xp^x , where x = 0, 1, 2, …n

It can be clearly observed that P(X = x) will be maximum if ⁶c_x will bw maximum.

∴⁶c_x = ⁶c₆ = 1

⁶c₁ = ⁶c₅ = 6

⁶c₂ = ⁶c₄ = 15

⁶c₃ = 20

Hence we can clearly see that ⁶c₃ is maximum.

∴for x = 3, P(X = x) is maximum.

Hence, proved that the most likely outcome is x = 3.

In this question, we have the repeated correct answer guessing form the given multiple choice questions are Bernouli trials

Let us now assume, X represents the number of correct answers by guessing in the multiple choice set

Now, probability of getting a correct answer,

Thus,

Clearly, we have X is a binomial distribution where n = 5 and

∴

Hence, probability of guessing more than 4 correct answer

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

(a) In this question, let X represents the number of prizes winning in 50 lotteries and the trials are Bernoulli trials

Here clearly, we have X is a binomial distribution where n = 50 and

Thus,

∴

Hence, probability of winning in lottery atleast once

= 1 – P (X <1)

= 1 – P (X = 0)

(b) Probability of winning in lottery exactly once

(c) Probability of winning in lottery atleast twice

= 1 – P (X < 2)

=

In this question, let us assume X represent the number of times of getting 5 in 7 throws of the die

Also, the repeated tossing of a die are the Bernoulli trials

Thus, probability of getting 5 in a single throw,

And,

Clearly, we have X has the binomial distribution where n = 7 and

∴

Hence, probability of getting 5 exactly twice in a die = P (X = 2)

In this question, let us assume X represent the number of times of getting sixes in 6 throws of a die

Also, the repeated tossing of die selection are the Bernoulli trials

Thus, probability of getting six in a single throw of die,

Clearly, we have X has the binomial distribution where n = 6 and

And,

∴

_{Hence, probability of throwing at most 2 sixes = P (X ≤2)}

_{= P (X = 0) + p (X = 1) + P (X = 2)}

In this question, let us assume X represent the number of times selecting defected articles in a random sample space of given 12 articles

Also, the repeated articles in a random sample space are the Bernoulli trials

Clearly, we have X has the binomial distribution where n = 12 and

And,

∴

Hence, probability of selecting 9 defective articles

In this question, let us assume X represent the number of times selecting defected bulbs in a random sample of given 5 bulbs

Also, the repeated selection of defective bulbs from a box are the Bernoulli trials

Clearly, we have X has the binomial distribution where n = 5 and

And,

∴

_{Hence, probability that none bulb is defective = P (X = 0)}

∴ option C is correct

In this question, let us assume X represent the number of students out of 5 who are swimmers

Also, the repeated selection of students who are swimmers are the Bernoulli trials

Thus, probability of students who are not swimmers

Clearly, we have X has the binomial distribution where n = 5

And,

∴

_{Hence, probability that four students are swimmers = P (X = 4)}

∴ option A is correct

It is given in the question that,

A and B are two events such that

We have,

∴

Hence,

= 1

(ii) We have,

∴

= 0

(i) According to the situation, if the couple has two children then the sample space is:

S = {(b, b), (b, g), (g, b), (g, g)}

Let us assume A denote the event of both children having male and B denote the event of having at least one of the male children

Thus, we have:

Hence,

(ii) Let us now assume C denote the event having both children females and D denote the event of having elder child is female

∴ C = {(g, g)}

And, D = {(g, b), (g, g)}

Hence,

It is given in the question that,

5% of men and 0.25% of women have grey hair

∴ Total % of people having grey hair = 5 + 0.25

= 5.25 %

Hence, Probability of having a selected person male having grey hair,

It is given in the question that,

90% of the people are right handed

Let p denotes the probability of people that are right handed and q denotes the probability of people that are left handed

∴

And,

Now, by using the binomial distribution probability of having more than 6 right handed people can be given as:

Hence, the probability of having more than 6 right handed people:

= 1 – P (More than 6 people are right handed)

=

(i) It is given in the question that,

Total number of balls in the urn = 25

Number of balls bearing mark ‘X’ = 10

Number of balls bearing mark ‘Y’ = 15

Let p denotes the probability of balls bearing mark ‘X’ and q denotes the probability of balls bearing mark ‘Y’

∴

And,

Now, 6 balls are drawn with replacement. Hence, the number of trials are Bernoulli triangle.

Let us assume, Z be the random variable that represents the number of balls bearing ‘Y’ mark in the trials

∴ Z has a binomial distribution where n = 6 and

Hence, P (All balls will bear mark ‘X’) = P (Z = 0)

(ii) Probability (Not more than 2 will bear ‘Y’ mark)

= P (Z = 0) + P (Z = 1) + P (Z = 2)

=

(iii) Now, Probability (At least one ball will bear ‘Y’ mark) = P (Z ≥ 1)

= 1 – P (Z = 0)

=

(iv) Probability (Having equal number of balls with ‘X’ mark and ‘Y’ mark) = P (Z = 3)

=

=

In this question,

Let us assume p be the probability of player that will clear the hurdle while q be the probability of player that will knock down the hurdle

∴

And,

Let us also assume X be the random variable that represents the number of times the player will knock down the hurdle

∴ By binomial distribution,

Hence, probability (players knocking down less than 2 hurdles) = P (X < 2)

= P (X = 0) + P (X = 1)

=

=

=

From the condition given in the question, it is clear that:

Probability of getting a six in a throw of die =

And, probability of not getting a six =

Let us assume,

Now, we have:

Probability that the 2 sixes come in the first five throws of the die

=

Also, Probability that the six come in the sixth throw

We know that, in a leap year there are total 366 days, 52 weeks and 2 days

Now, in 52 weeks there are total 52 Tuesdays

∴ Probability that the leap year will contain 53 Tuesdays is equal to the probability of remaining 2 days will be Tuesdays

Thus, the remaining two days can be:

(Monday and Tuesday), (Tuesday and Wednesday), (Wednesday and Thursday), (Thursday and Friday), (Friday and Saturday), (Saturday and Sunday) and (Sunday and Monday)

∴ Total Number of cases = 7

Cases in which Tuesday can come = 2

Hence, probability (leap year having 53 Tuesdays)

It is given in the question that,

Probability of failure = x

And, probability of success = 2x

∴ x + 2x = 1

3x = 1

∴

Let us now assume

Also, X be the random variable that represents the number of trials

Hence, by binomial distribution we have:

∴ Probability of having at least 4 successes

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

Let us assume that, man tosses the coin n times. Thus, n tosses are the Bernoulli trials

∴ Probability of getting head at the toss of the coin

Let us assume,

∴

It is given in the question that,

Probability of getting at least one head >

∴

1 – P (x = 0) > 0.9

(i)

Hence, the minimum value of n satisfying the given inequality = 4

∴ The man have to toss the coin 4 or more times

For the situation given in the equation, we have:

Probability of getting a six in a throw of a die

Also, probability of not getting a 6

Now, there are three cases from which the expected value of the amount which he wins can be calculated:

(i) First case is that, if he gets a six on his first through then the required probability will be

∴ Amount received by him = Rs. 1

(ii) Secondly, if he gets six on his second throw then the probability

∴ Amount received by him = - Rs. 1 + Rs. 1

= 0

(iii) Lastly, if he does not get six in first two throws and gets six in his third throw then the probability

∴ Amount received by him = - Rs. 1 – Rs. 1 + Rs. 1

= - 1

Hence, expected value that he can win

Let us assume R be the event of drawing the red marbles

Let us also assume E_A, E_B andE_C denote the boxes A, B and C respectively

It is given in the question that,

Total number of marbles = 40

Also, total number of red marbles = 15

∴

Probability of taking out the red marble from box A,

Also, probability of taking out the red marble from box B,

And, Probability of taking out the red marble from box C,

Let us assume, X denotes the events having a person heart attack

A₁ denote events having the selected person followed the course of yoga and meditation

And, A₂ denote the events having the person adopted the drug prescription

It is given in the question that,

P (X) = 0.40

And,

∴ Probability (The patient suffering from a heart attack and followed a course of meditation and yoga):

=

From the question, we have:

Total number of determinants of second order where the element being or 1 = (2)⁴

= 16

Now, we have the value of determinants is positive in following cases:

∴ Required probability = 3/16

(i) Let us assume the event which is failed by A is denoted by E_A

And, event which is failed by B is denoted by E_B

It is given in the question that,

Event failed by A, P (E_A) = 0.2

Event failed by both,

And, event failed by B alone

0.15 = P (E_B) – 0.15

∴ P (E_B) = 0.30

Hence,

= 0.5

(ii) We have,

Probability where A fails alone

Let us firstly assume, A₁ denote the events that a red ball is transferred from bag I to II

And, A₂ denote the event that a black ball is transferred from bag I to II

∴

And,

Let X be the event that the drawn ball is red

∴ when red ball is transferred from bag I to II,

And, when black ball is transferred from bag I to II,

Hence,

It is given in the question that,

A and B are two events where,

And, P (B|A) = 1

∴

∴

Hence, option A is correct

It is given in the question that,

P (A|B) > P (A)

∴

P (B|A) > P (B)

Hence, option C is correct

It is given in the question that,

A and B are any two events where,

P (A) + P (B) – P (A and B) = P (A)

∴

= 1

Hence, option B is correct