Question

NCERT Solutions Class 12 Maths Chapter 13 Probability is one of the best study materials for the students who are preparing for their board exams and also for competitive exams as well. Our NCERT Solutions has a complete discussion of all kinds of topics, from easy to difficult concepts. NCERT Solutions Class 12 is made in a pointwise method.

• Probability – probability means the possibility of happening an event. In the case of probability, the random event occurs according to some random event. Probability is introduced in the math to find out how likely the event is about to happen. The probability is calculated in the range of 0 to 1. It is the probability of a single event that tends to occur first when we know the total number of possible outcomes. The algebraic sum of the entire outcome in a sample space adds up to 1. Many events cannot be predicted with complete certainty.
• Conditional Probability – such outcome of which depends upon the outcome of the previous event. The conditional probability is calculated by multiplying the new probability with the preceding probability.
• Marginal probability – is the probability of those events which can be mentioned by unconditional probability. The marginal probability depends upon the occurrence of another event.
• Joint probability – when two events are happening simultaneously when these two events have the intersection of two or more events.  
• Multiplication Theorem on Probability – such events which occur at the same time with the event A and B are equal to the product of the probability of B happening. Conditional probability is the related occurrence of different events when another event occurs.
• Independent Events – in probability in which outcomes of an experiment are calculated with different types of events are called independent events. It is also a mutually exclusive event.

NCERT Solutions Class 12 Maths is made by the experts to gain a good clarity of all kinds of concepts.

Answer 1

NCERT Solutions Class 12 Maths Chapter 13 Probability

1. Given that E and F are events such that P (E) = 0.6, P (F) = 0.3 and P (E ∩ F) = 0.2, find P (E|F) and P (F|E)

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

We know that by the definition of conditional probability,

2. Compute P (A|B), if P (B) = 0.5 and P (A ∩ B) = 0.32

Answer:  P(B) = 0.5 and P(A ∩ B) = 0.32

We know that by the definition of conditional probability,

3. If P (A) = 0.8, P (B) = 0.5 and P (B|A) = 0.4, find

(i) P (A ∩ B)

(ii) P (A|B)

(iii) P (A ∪ B)

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

(i)  We know that by the definition of conditional probability,

(ii) We know that by the definition of conditional probability,

(iii) Now,  ∵ P(A U B) = P(A) + P(B) - P(A ∩ B)

Substituting the values we get

⇒ P(A U B) = 0.8 + 0.5 - 0.32 = 1.3 - 0.32

⇒ P(A U B) = 0.98

4. Evaluate P (A ∪ B), if 2P (A) = P (B) = 5/13 and P (A|B) = 2/5.

Answer:

We know that by the definition of conditional probability,

5. If P (A) = 6/11, P (B) = 5/11 and P (A ∪ B) = 7/11, find

(i) P (A∩B)

(ii) P (A|B)

(iii) P (B|A)

Answer:  Given P(A) = 6/11, P(B) = 5/11, P(A U B) = 7/11

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

(ii) Now, by definition of conditional probability,

(iii) Again, by definition of conditional probability,

6. A coin is tossed three times, where

(i) E : head on third toss, F : heads on first two tosses

(ii) E : at least two heads, F : at most two heads

(iii) E : at most two tails, F : at least one tail

Answer: The sample space of the given experiment will be:

Now, we know that by definition of conditional probability,

7. Two coins are tossed once, where

(i) E: tail appears on one coin, F: one coin shows head

(ii) E: no tail appears, F: no head appears

Answer: The sample space of the given experiment is S = {HH, HT, TH, TT}

Now, we know that by definition of conditional probability

(ii) Here, E: no tail appears

And F: no head appears

⇒ E = {HH} and F = {TT}

⇒ E ∩ F = ϕ

So, P(E) = 1/4, P(F) = 1/4, P(E ∩ F) = 0/4 = 0

Now, we know that by  definition of conditional probability,

Answer 2

8. A die is thrown three times, E: 4 appears on the third toss, F: 6 and 5 appears respectively on first two tosses.

Answer : 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 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)}

P(E) = \(\frac{36}{216}\), P(F) = \(\frac6{216}\), P(E ∩ F) = \(\frac1{216}\)

Now, we know that by definition of conditional probability,

Now by substituting the values we get

9. Mother, father and son line up at random for a family picture E: son on one end, F: father in middle

Answer :

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}

Now, we know that by definition of conditional probability,

Now, by substituting the values we get

10. A black and a red dice are rolled.

(a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.

(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

Answer :

Let B denote black coloured die and R denote red coloured 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)}

An B = {(B5, R5), (B5, R6)}

So, P(A) = \(\frac6{36}\) = \(\frac16\), P(B) = \(\frac6{36}\) = \(\frac16\), P(A ∩ B) = \(\frac2{36}=\frac1{18}\)

Now, we know that by definition of conditional probability,

Now, by substituting the values we get

(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)}

P(A) = 5/36, P(B) = 18/36 = 1/2, P(A ∩ B) = 2/36 = 1/18

Now, we know that

By definition of conditional probability,

Now by substituting the values we get

11. A fair die is rolled. Consider events E = {1,3,5}, F = {2,3} and G = {2,3,4,5} Find

(i) P (E|F) and P (F|E)

(ii) P (E|G) and P (G|E)

(iii) P ((E ∪ F)|G) and P ((E ∩ F)|G)

Answer :

The sample space for the given experiment is S = {1, 2, 3, 4, 5, 6} 

Here, E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5} --(i)

⇒ P(E) = 3/6 = 1/2, P(F) = 2/6 = 1/3, P(G) = 4/6 = 2/3 --(ii)

Now, E ∩ F = {3}, F ∩ G = {2, 3}, E ∩ G = {3, 5}----(iii)

⇒ P(E ∩ F) = 1/6, P(F ∩ G) = 2/6 = 1/3, P(E ∩ G) = 2/6 = 1/3----(iv)

(i) Now, we know that by definition of conditional probability,

(ii) Now, we know that by definition of conditional probability,

12. Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl?

Answer :

Let B denote boy and G denote girl.

Then, the sample space of the given experiment is S = {GG, GB, BG, BB}

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

⇒ E = {GG}

⇒ P(E) = 1/4

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

⇒ F = {GG, BG}

⇒ P(F) = 2/4 = 1/4....(i)

Now, E ∩ F = {GG}

⇒ P(E ∩ F) = 1/4....(ii)

Now, we know that by definition of conditional probability,

(ii) Let H be the event that 'at least one is girl'.

Now, we know that by definition of conditional probability,

Answer 3

13. An instructor has a question bank consisting of 300 easy True / False questions, 200 difficult True / False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?

Answer : 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'.

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, P(E ∩ F) = 500/1400 = 5/14---(ii)

Now, we know that by definition of conditional probability,

14. Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.

Answer : The sample of the given experiment is given below

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)}

⇒ P(E) = 3/36 = 1/12, P(F) = 30/36 = 5/6, P(E ∩ F) = 2/36 = 1/18......(i)

Now, we know that by definition of conditional probability,

15. Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.

Answer :  The experiment is explained below in the tree diagram:

The sample space of the given experiment is given below

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,

16. If P (A) = 1/2, P (B) = 0, then P (A|B) is

A. 0

B. 1/2

C. not defined

D. 1

Answer : (C) not defined

Explanation:

Now, we know that by definition of conditional probability,

which is not defined.

17. If A and B are events such that P (A|B) = P (B|A), then

A. A ⊂ B but A ≠ B

B. A = B

C. A ∩ B = φ

D. P (A) = P (B)

Answer : (D) P (A) = P (B)

Given: P(A|B) = P(B|A)....(i)

Now, we know that by definition of conditional probability,

Answer 4

18. If P (A) = 3/5 and P (B) = 1/5, find P (A ∩ B) if A and B are independent events.

Answer :

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

As A and B are independent events.

⇒ P (A ∩ B) = P (A).P (B)

= 3/5 × 1/5 = 3/25

19. Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

Answer :

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) = 26/52 = 1/2

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 that is the conditional probability of B given that A has already occurred.

Now, P (B/A) = P (black card in second draw) = 25/51

Thus the probability that both the cards are black

⇒ P (A ∩ B) = 1/2 × 25/51 = 25/102

Hence, the probability that both the cards are black = 25/102.

20. A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.

Answer :

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) = 12/15

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 that is the conditional probability of B given that A has already occurred.

Now, P (B/A) = P (good orange in second draw) = 11/14

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 that is the conditional probability of C given that A and B has already occurred.

Now, P (C/AB) = P (good orange in third draw) = 10/13

Thus the probability that all the oranges are good

⇒ P (A ∩ B ∩ C) = 12/15 × 11/14 × 10/13 = 44/91

Hence, the probability that a box will be approved for sale = 44/91

21. A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not.

Answer :

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) = 6/12 =1/2

Now, Let B be the event 3 on the die

⇒ B = {(H, 3), (T, 3)}

⇒ P (B) = 2/12 = 1/6

As, A ∩ B = {(H, 3)}

⇒ P (A ∩ B) = 1/12 …… (1)

And P (A). P (B) =1/2 × 1/6 = 1/12 …… (2)

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

Therefore, A and B are independent events.

22. A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, ‘the number is even,’ and B be the event, ‘the number is red’. Are A and B independent?

Answer :

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}

⇒ P (A) = 3/6 = 1/2

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

⇒ B = {1, 2, 3}

⇒ P (B) = 3/6 = 1/2

As, A ∩ B = {2}

⇒ P (A ∩ B) = 1/6 …….. (1)

And P (A). P (B) = 1/2 × 1/2 =1/4 ….. (2)

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

Therefore, A and B are not independent events.

23. Let E and F be events with P (E) = 3/5, P (F) = 3/10 and P (E ∩ F) = 1/5. Are E and F independent?

Answer :

Given P (E) = 3/5, P (F) = 3/10 and P (E ∩ F) = 1/5

P (E). P (F) = 3/5 × 3/10 = 9/50 ≠ 1/5

⇒ P (E ∩ F) ≠ P (E). P (F)

Therefore, E and F are not independent events.

24. Given that the events A and B are such that P (A) = 1/2, P (A ∪ B) = 3/5 and P (B) = p. Find p if they are (i) mutually exclusive (ii) independent.

Answer :

Given P (A) = 1/2, P (A ∪ B) = 1/5 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)

⇒ 3/5 = 1/2 + p -0

⇒ P = 3/5 – 1/2 = 1/10

(ii) Independent

When A and B are independent.

⇒ P (A ∩ B) = P (A). P (B)

⇒ P (A ∩ B) = 1/2 p

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

⇒ 3/5 = 1/2 + 2 – p/2

⇒ p/2 = 3/5 – 1/2

⇒ p = 2 × 1/10 = 1/5

25. Let A and B be independent events with P (A) = 0.3 and P (B) = 0.4. Find

(i) P (A ∩ B)

(ii) P (A ∪ B)

(iii) P (A|B)

(iv) P (B|A)

Answer :

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

26. If A and B are two events such that P (A) = 1/4 , P (B) = 1/2 and P (A ∩ B) = 1/8, find P (not A and not B).

Answer :

Given P (A) = 1/4, P (B) = 1/2 and P (A ∩ B) = 1/8

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)]

Answer 5

27. Events A and B are such that P (A) = 1/2, P (B) = 7/12 and P (not A or not B) = 1/4. State whether A and B are independent?

Answer :

Given P (A) = 1/2, P (B) =7/12 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) = 1/2 × 7/12 = 7/24 …. (2)

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

Therefore, A and B are not independent events.

28. Given two independent events A and B such that P (A) = 0.3, P(B) = 0.6. Find

(i) P (A and B)

(ii) P (A and not B)

(iii) P (A or B)

(iv) P (neither A nor B)

Answer :

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

29. A die is tossed thrice. Find the probability of getting an odd number at least once.

Answer :

Given a die is tossed thrice.

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

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

⇒ P (A) = 3/6 =1/2.

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

⇒ P (B) =3/6 = 1/2.

Now, probability of getting an even number in three times = 1/2 × 1/2 × 1/2 = 1/8

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

= 1 – 1/8

∴ Probability of getting an odd number at least once = 7/8

30. Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that

(i) both balls are red.

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

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

Answer :

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 = 8/18 = 4/9

As the ball is replaced after first throw,

Hence, Probability of getting a red ball in second draw = 8/18 = 4/9

Now, Probability of getting both balls red = 4/9 × 4/9 = 16/81

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

Probability of getting a black ball in first draw = 10/18 = 5/9

As the ball is replaced after first throw,

Hence, Probability of getting a red ball in second draw = 8/18 = 4/9

Now, Probability of getting first ball is black and second is red = 5/9 × 5/9 = 20/81

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

Probability of getting a black ball in first draw = 10/18 = 5/9

As the ball is replaced after first throw,

Hence, Probability of getting a red ball in second draw = 8/18 = 4/9

Now, Probability of getting first ball is black and second is red = 5/9 × 4/9 = 20/81

Probability of getting a red ball in first draw = 8/18 = 4/9

As the ball is replaced after first throw,

Hence, Probability of getting a black ball in second draw = 10/18 = 5/9

Now, Probability of getting first ball is red and second is black = 5/9 × 4/9 = 20/81

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

= 20/81 + 20 /81 = 40/81

31. Probability of solving specific problem independently by A and B are 1/2 and 1/3 respectively. If both try to solve the problem independently, find the probability that

(i) The problem is solved

(ii) Exactly one of them solves the problem.

Answer :

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) = 1/2 × 1/3 = 1/6

P (A^’) = 1 – P (A) = 1 – 1/2 = 1/2

P (B^’) = 1 – P (B) = 1 – 1/3 = 2/3

(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) = 1/2 + 1/3 – 1/6 = 4/6

⇒ P (A ∪ B) = 2/3

(ii) Exactly one of them solves the problem

That is either problem is solved by A but not by B or vice versa

That is P (A).P (B^’) + P (A^’).P (B)

= 1/2  (2/3) +1/2 (1/3)

= 1/3 + 1/6 = 3/6

⇒ P (A).P (B^’) + P (A^’).P (B) = 1/2

Answer 6

32. One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent?

(i) E: ‘the card drawn is a spade’ F: ‘the card drawn is an ace’

(ii) E: ‘the card drawn is black’ F: ‘the card drawn is a king’

(iii) E: ‘the card drawn is a king or queen’ F: ‘the card drawn is a queen or jack’.

Answer :

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 = 13/52 = 1/4

P (F) = the card drawn is an ace = 4/52 = 1/13

P (E ∩ F) = the card drawn is a spade and ace both = 1/52….. (1)

And P (E). P (F)

= 1/4 × 1/13 = 1/52…. (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 = 26/52 = 1/2

P (F) = the card drawn is a king = 4/52 = 1/13

P (E ∩ F) = the card drawn is a black and king both = 2/52 = 1/26…. (1)

And P (E). P (F)

= 1/2 × 1/13 = 1/26…. (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 = 8/52 = 2/13

P (F) = the card drawn is either queen or jack = 8/52 = 2/13

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 = 4/52 = 1/13 … (1)

And P (E). P (F)

= 2/13 × 2/13 = 4/169…. (2)

From (1) and (2)

⇒ P (E ∩ F) ≠ P (E). P (F)

Hence, E and F are not independent events.

32. In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random.

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

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

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

Answer :

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= 60/100 = 3/5

P (E) = Probability of students who read English newspaper = 40/100 = 2/5

P (H ∩ E) = Probability of students who read Hindi and English both newspaper = 20/100 = 1/5

(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)  = \(\frac{P(H \cap E)}{P(H)}\)

⇒ P(E|H) = \(\cfrac{\frac12}{\frac25}=\frac15\times\frac53\)

⇒ P (E|H) = 1/3

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

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

As we know P (H|E) = \(\frac{P(H\cap E)}{P(E)}\) 

⇒ P (H|E) = \(\cfrac{\frac15}{\frac25}=\frac15\times\frac52\)

⇒ P(H|E) = 1/2

32. The probability of obtaining an even prime number on each die, when a pair of dice is rolled is

A. 0

B. 1/3

C. 1/12

D. 1/36

Answer : D. 1/36

Explanation:

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}

⇒ P € = 1/36

33. Two events A and B will be independent, if

(A) A and B are mutually exclusive

(B) P (A′B′) = [1 – P (A)] [1 – P (B)] (C) P (A) = P (B)

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

Answer : (B) P (A′B′) = [1 – P (A)] [1 – P (B)]

Explanation:

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

34. An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

Answer :

Given 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) = 5/10 = 1/2

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) = 7/12

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

⇒ P (probability of drawing a black ball) = 5/10 = 1/2

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) = 5/12

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

= (1/2 x 7/12) + (1/2 x 5/12) = 1/2 (7/12 + 5/12) = 1/2 x 1 = 1/2

Answer 7

35. A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Answer :

Let E₁ be the event of choosing the bag I, E₂ be the event of choosing the 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) = 4/8 = 1/2

And P (A|E₂) = P (drawing a red ball from bag II) = 2/8 = 1/4

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:

36. Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostlier?

Answer :

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.

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:

substituting the values we get

37. In answering a question on a multiple choice test, a student either knows the answer or guesses. Let 3/4 be the probability that he knows the answer and 1/4 be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability 1/4. What is the probability that the student knows the answer given that he answered it correctly?

Answer :

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 P (E₁) = 3/4

And P (E₂) = 1/4

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

And P (A|E₂) = P (correct answer given that he guesses) = 1/4

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

By using Bayes’ theorem, we have:

38. A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?

Answer :

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 P (E₁) = 0.1% = 0.1/100 = 0.001 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

39. There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?

Answer :

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) = 75% = 75/100 = 3/4

And P (A|E₃) = P (coin shows head given that the coin is unbiased) = 1/2

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

By substituting the values we get

40. An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

Answer :

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 P (E₁) = 2000/12000 = 1/6

P (E₂) = 4000/12000 = 1/3

P (E₃) = 6000/12000 =1/2

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) = 0.01 = 1/100

And P (A|E₂) = P (accident of a car driver) = 0.03 = 3/100

And P (A|E₃) = P (accident of a truck driver) = 0.15 = 15/100 = 3/20

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

Now by substituting the values we get

Answer 8

41. A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?

Answer :

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 P (E₁) = 60% = 60/100 = 3/5

P (E₁) = 40% = 40/100 = 2/5

Also P (X|E₁) = P (item is defective given that it is produced by machine A) = 2% = 2/100 = 1/50

And P (X|E₂) = P (item is defective given that it is produced by machine B) = 1% = 1/100

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

By substituting the values we get

42. Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.

Answer :

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

Now by substituting the value we get

P(E₂|A) = 2/9

43. Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?

Answer : 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 P (E₁) = 2/6 = 1/3

P (E₂) = 4/6 = 2/3

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) = 3/8

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

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

Now by substituting the values we get

P(E₂|A) = 8/11

44. A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?

Answer :

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 P (E₁) = 50% = 50/100 = 1/2

P (E₂) = 30% = 30/100 = 3/10

P (E₃) = 20% = 20/100 = 1/5

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) = 1% = 1/100

And P (X|E₂) = P (defective item produced by B) = 5% = 5/100

And P (X|E₃) = P (defective item produced by C) = 7% = 7/100

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

Now by substituting the values we get

P(E₁|X) = 5/34

45. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

Answer :

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 P (E₁) = 13/52 and P (E₂) = 39/52

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 (E₁|A).

By using Bayes' theorem, we have:

Now by substituting the values we get

Answer 9

46. Probability that A speaks truth is 4/5. A coin is tossed. A reports that a head appears. The probability that actually there was head is

A. 4/5

B. 1/2

C. 1/5

D. 2/5

Answer : A. 4/5

Explanation:

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, P (E₁) = 4/5

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) = 1 – 4/5 = 1/5

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

Now substituting the values we get

47. If A and B are two events such that A ⊂ B and P(B) ≠ 0, then which of the following is correct?

A. P (A| B) = P (B)/ P (A)

B. P(A|B) < P(A)

C. P(A|B) ≥ P(A)

D. None of these

Answer : (C) P (A|B) ≥ P (A)

Explanation:

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

48. State which of the following are not the probability distributions of a random variable. Give reasons for your answer.

(i) 

X	0	1	2
P(X)	0.4	0.4	0.2

Answer : Here we have 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.

(ii) 

X	0	1	2	3	4
P(X)	0.1	0.5	0.2	-0.1	0.3

Answer : Here we have 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.

(iii) 

Y	-1	0	1
P(Y)	0.6	0.1	0.2

Answer : Here we have 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.

(iv) 

Z	3	2	1	0	-1
P(Z)	0.3	0.2	0.4	0.1	0.05

Here we have table with value 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 \(\neq\) 1

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

48. An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X? Is X a random variable?

Answer :

Given 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.

49. Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. What are possible values of X?

Answer :

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.

Answer 10

50. Find the probability distribution of

(i) number of heads in two tosses of a coin.

(ii) Number of tails in the simultaneous tosses of three coins.

(iii) Number of heads in four tosses of a coin.

Answer:

(i) number of heads in two tosses of a coin.

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,

X	0	1	2
P (X)	1/4	1/2	1/4

 (ii) Number of tails in the simultaneous tosses of three coins.

Given 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) = 1/8

P (X = 1) = P (HHT) + P (HTH) + P (THH) = 1/8 + 1/8 + 1/8 = 3/8

P (X = 2) = P (HTT) + P (THT) + P (TTH) = 1/8 + 1/8 + 1/8 = 3/8

P (X = 3) = P (TTT) = 1/8

Hence, the required probability distribution is,

X	0	1	2	3
P (X)	1/8	3/8	3/8	1/8

(iii) Number of heads in four tosses of a coin.

Given 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) = 1/16

P (X = 1) = P (HTTT) + P (TTTH) + P (THTT) + P (TTHT) = 1/16 + 1/16 + 1/16 + 1/16 = ¼

P(X = 2) = P (HHTT) + P (THHT) + P (TTHH) + P (THTH) + P (HTHT) + P(HTTH)= 1 /16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 = 6/16 = 3/8

P(X = 3) = P (THHH) + P (HHHT) + P (HTHH) + P (HHTH) = 1/16 + 1/16 + 1/16 + 1/16 = ¼

P(X = 4) = P (HHHH) = 1/16

Hence, the required probability distribution is,

X	0	1	2	3	4
P (X)	1/16	1/4	3/8	1/4	1/16

51. Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as

(i) number greater than 4

(ii) six appears on at least one die

Answer:

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) = 4/6 × 4/6 = 4/9

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

= (4/6 × 2/6) + (2/6 × 4/6) = 4/9

P (X = 2) = P (number ≥ 4 in both tosses) = 2/6 × 2/6 = 1/9

Hence, the required probability distribution is,

X	0	1	2
P (X)	4/9	4/9	1/9

(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) = 5/6 × 5/6 = 25/36

P (X = 1) = P (six appears at least once of the die) = (1/6 × 5/6) + (5/6 × 1/6) = 10/36 = 5/18

Hence, the required probability distribution is,

X	0	1
P (X)	25/36	5/18

52. From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs. 

Answer:

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.

Hence, the required probability distribution is,

X	0	1	2	3	4
P (X)	256/625	256/625	96/625	16/625	1/625