# CLASS 12 MATHS CHAPTER 13-PROBABILITY

### Solve The Following Questions.

Question1. 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).

Solution :
It is given that P(E) = 0.6, P(F) = 0.3, and P(E ∩ F) = 0.2

Question2. Compute P(A|B), if P(B) = 0.5 and P (A ∩ B) = 0.32

Solution :
Given: P (B) = 0.5, P (A ∩  B) = 0.32

Question3.

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)

Solution : Question4. Evaluate P (A ∪ B), if 2P (A) = P (B) =5/13 and P(A|B) =2/5

Solution : Question 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)

Solution :
It is given that,P(A)=6/11, P(B) =5/11 and P(A ∪ B) =7/11

Question6. Determine: A coin is tossed three times.

(i) E : heads 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.

Solution :

If a coin is tossed three times, then the sample space S is

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

It can be seen that the sample space has 8 elements.

(i) E = {HHH, HTH, THH, TTH}

F = {HHH, HHT}

∴ E ∩ F = {HHH}

(ii) E = {HHH, HHT, HTH, THH}

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

∴ E ∩ F = {HHT, HTH, THH}

(iii) E = {HHH, HHT, HTT, HTH, THH, THT, TTH}

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

Question7. Determine :Two coins are tossed once.

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

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

Solution :

If two coins are tossed once, then the sample space S is

S = {HH, HT, TH, TT}

(i) E = {HT, TH}

F = {HT, TH}

(ii) E = {HH}

F = {TT}

∴ E ∩ F = Φ

P (F) = 1 and P (E ∩ F) = 0

∴ P(E|F) = Question8. Determine :E A dice is thrown three times.

E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses.

Solution :
Since a dice has six faces. Therefore E = 6 x 6 x 6 = 216

E = (1, 2, 3, 4, 5, 6) x (1, 2, 3, 4, 5, 6) x (4)

F = (6) x (5) x (1, 2, 3, 4, 5, 6)

Question9. Determine : Mother, father and son line up at random for a family picture.

E : Son on one end, F : Father in middle.

Solution :

If mother (M), father (F), and son (S) line up for the family picture, then the sample space will be

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

⇒ E = {MFS, FMS, SMF, SFM}

F = {MFS, SFM}

∴ E ∩ F = {MFS, SFM}

E : Son on one end

Question10. A black and a red die 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.

Solution :

Let the first observation be from the black die and second from the red die.

When two dice (one black and another red) are rolled, the sample space S has 6 × 6 = 36 number of elements.

(a) Let A: Obtaining a sum greater than 9

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

B: Black die results in a 5.

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

∴ A ∩ B = {(5, 5), (5, 6)}

The conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5, is given by P (A|B).

(b) E: Sum of the observations is 8.

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

F: Red die resulted in a number less than 4.

The conditional probability of obtaining the sum equal to 8, given that the red die resulted in a number less than 4, is given by P (E|F).

Question11. 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)

(ii) P ((E ∪ F)|G) and P ((E ∩ G)|G)

Solution :

When a fair die is rolled, the sample space S will be

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

It is given that E = {1, 3, 5}, F = {2, 3}, and G = {2, 3, 4, 5}

(iii) E ∪ F = {1, 2, 3, 5}

(E ∪ F) ∩ G = {1, 2, 3, 5} ∩{2, 3, 4, 5} = {2, 3, 5}

E ∩ F = {3}

(E ∩ F) ∩ G = {3}∩{2, 3, 4, 5} = {3}

Question12. Assume that each child born 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?

Solution :

Let b and g represent the boy and the girl child respectively. If a family has two children, the sample space will be

S = {(bb), (bg), (gb), (g, g)}

Let A be the event that both children are girls.

∴ A = {(g,g)}

(i) Let B be the event that the youngest child is a girl.

The conditional probability that both are girls, given that the youngest child is a girl, is given by P (A|B).

Therefore, the required probability is 1/2.

(ii) Let C be the event that at least one child is a girl.

The conditional probability that both are girls, given that at least one child is a girl, is given by P(A|C).

Question13. An instructor has a test 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 test bank, what is the probability that it will be an easy question given that it is a multiple choice question?

Solution :

The given data can be tabulated as

Let us denote E = easy questions, M = multiple choice questions, D = difficult questions, and T = True/False questions

Total number of questions = 1400

Total number of multiple choice questions = 900

Therefore, probability of selecting an easy multiple choice question is

P (E ∩ M) = 500/1400 = 5/14

Probability of selecting a multiple choice question, P (M), is

900/1400 = 9/14

P (E|M) represents the probability that a randomly selected question will be an easy question, given that it is a multiple choice question.

∴ P(E|M) = P(E ∩ M)/P(M) = 5/14/9/14 = 5/9

Therefore, the required probability is 5/9.

Question14. 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’.

Solution :

When dice is thrown, number of observations in the sample space = 6 × 6 = 36

Let A be the event that the sum of the numbers on the dice is 4 and B be the event that the two numbers appearing on throwing the two dice are different.
∴ A = {(1, 3), (2, 2), (3, 1)
S =  (1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1)

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

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

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

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

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

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

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

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

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

(3, 6), (4, 6), (5, 6)}

Let P (A|B) represent the probability that the sum of the numbers on the dice is 4, given that the two numbers appearing on throwing the two dice are different.

Therefore, the required probability is 1/15.

Question15. 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”.

Solution :

The outcomes of the given experiment can be represented by the following tree diagram.

The sample space of the experiment is,

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

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

Let A be the event that the coin shows a tail and B be the event that at least one die shows 3.

Probability of the event that the coin shows a tail, given that at least one die shows 3, is given by P(A|B).

Therefore,

Question16. If P (A) = 1/2 P (B) = 0, then P(A|B) is:

(A) 0

(B) 1/2

(C) not defined

(D) 1

Solution :
P (A) = 1/2  P(B) = 0

Therefore, P (A|B) is not defined.

Therefore, option (C) is correct.

Question17. 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)

Solution :

It is given that, P(A|B) = P(B|A)

⇒ P (A) = P (B)

Therefore, option (D) is correct.

### Solve The Following Questions.

Question1. If P (A) = 3/5  and P (B) = 1/5 find P (A ∩ B) if A and B are independent events.

Solution :

It is given that 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

Question2. 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.

Solution :

There are 26 black cards in a deck of 52 cards.

Let P (A) be the probability of getting a black card in the first draw.

P(A) = 26/52 = 1/2

Let P (B) be the probability of getting a black card on the second draw.

Since the card is not replaced,

P(B) = 25/51

Thus, probability of getting both the cards black = 1/2 x25/51 = 25/102

Question3. 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.

Solution :

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

Therefore, probability that first drawn orange is good, P (A) = 12/15

The oranges are not replaced.

Therefore, probability of getting second orange good, P (B) = 11/14

Similarly, probability of getting third orange good, P(C) = 10/13

The box is approved for sale, if all the three oranges are good.

Thus, probability of getting all the oranges good = 12/15 x 11/14 x 10/13 = 44/91

Therefore, the probability that the box is approved for sale is 44/91.

Question4. 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.

Solution :

If a fair coin and an unbiased die are tossed, then the sample space S is given by,

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: Head appears on the coin

Therefore, A and B are independent events.

Question5. A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event ‘number is even’ and B be the event ‘number is red’. Are A and B independent?

Solution :

When a die is thrown, the sample space (S) is

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

Let A: the number is even = {2, 4, 6}

P (A) = 3/6 = 1/2

B: the number is red = {1, 2, 3}

P (B) = 3/6 = 1/2

∴ A ∩ B = {2}

Therefore, A and B are not independent.

Question6. Let E and F be events with  Are E and F independent?

Solution : Therefore, E and F are not independent events.

Question7. 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.

Solution :

It is given that P(A) = 1/2 P (A ∩ B) = 3/5  and P (B) = p

(i) When A and B are mutually exclusive, A ∩ B = Φ

∴ P (A ∩ B) = 0

It is known that, P(A U B) = P(A) + P(B) – P(A ∩ B)

(ii) When A and B are independent, P(A ∩ B) = P(A) .P(B) = 1/2p

It is known that, P(A U B) = P(A) + P(B) – P(A ∩ B)

Question8. Let A and B independent events, P (A) = 0.3 and P (B) = 0.4. Find:

(A) P (A ∩ B)

(B) P (A U B)

(C) P (A|B)

(D) P (B|A)

Solution :
P (A) = 0.3, P (B) = 0.4

A and B are independent events.

(i) P (A ∩ B) P (A). P (B) = 0.3 x 0.4 = 0.12

(ii) P (A U B) = P (A) + P (B) – P (A). P (B) = 0.03 + 0.4 – 0.3 x 0.4 = 0.7 – 0.12 = 0.58

(iii) P (A|B) = P (A ∩ B)/P(B) = P (A|B) = P (A) = 0.3

(iv) P(B|A) = P (A ∩ B)/P(A) = P(B|A) = P (B) = 0.4

Question9. 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).

Solution :

It is given that, P (A) = 1/4 and P (A ∩ B) = 1/8

P(not on A and not on B) =P(A’ ∩  B’)

P(not on A and not on B) P(A U B)’ [A’ ∩ B’ = (A U B)’]

Question10. 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.

Solution :
It is given that  P (A) = 1/2 P (B) = 7/12 and P (not A or not B) = 1/4

Question11. Given two independent events A and B such that P (A) = 0.3, P (B) = 0.6. Find:

(A) P (A and B)

(B) P (A and not B)

(C) P (A or B)

(D) P (neither A nor B)

Solution :
P (A) = 0.3, P (B) = 0.6

A and B are independent events.

(i) P (A and B) = P (A). P (B) = 0.3 x 0.6 = 0.18

(ii) 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) + P (B) – P (A and B) = 0.3 + 0.6 – 0.18 = 0.9 – 0.18 = 0.72

(iv) P (neither A nor B) =  P(A’ ∩ B’) = (P(A U B’)) 1 – P (A U B) = 1 – 0.72 = 0.28

Question12. A die is tossed thrice. Find the probability of getting an odd number at least once.

Solution :

Probability of getting an odd number in a single throw of a die = 3/6 = 1/2

Similarly, probability of getting an even number =3/6 = 1/2

Probability of getting an even number three times = 1/2 x 1/2 x 1/2 = 1/8

Therefore, probability of getting an odd number at least once

= 1 − Probability of getting an odd number in none of the throws

= 1 − Probability of getting an even number thrice

= 1- 1/8 = 7/8

Question13. 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.

Solution :

Total number of balls = 18

Number of red balls = 8

Number of black balls = 10

(i) Probability of getting a red ball in the first draw = 8/18 = 4/9

The ball is replaced after the first draw.

∴ Probability of getting a red ball in the second draw = 8/18 = 4/9

Therefore, probability of getting both the balls red = 4/9 x 4/9 = 16/81

(ii) Probability of getting first ball black =10/18 = 5/9

The ball is replaced after the first draw.

Probability of getting second ball as red = 8/18 = 4/9

Therefore, probability of getting first ball as black and second ball as red =5/9 x 4/9 = 20/81

(iii) Probability of getting first ball as red =8/18 = 4/9

The ball is replaced after the first draw.

Probability of getting second ball as black =10/18 = 5/9

Therefore, probability of getting first ball as black and second ball as red =  4/9 x 5/9 = 20/81

Therefore, probability that one of them is black and other is red

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

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

Question14. 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.

Solution :

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

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

Since the problem is solved independently by A and B,

∴ P(AB) = P(A).P(B) = 1/2 x 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)Probability that the problem is solved = P (A ∪ B)

= P (A) + P (B) − P (AB) = 1/2 + 1/3 -1/6 = 4/6 = 2/3

(ii) Probability that exactly one of them solves the problem is given by, P(A).P(B’) + P(B).P(A’)

= 1/2 x 2/3 + 1/2 x 1/3 = 1/3 + 1/6 = 1/2

Question15. 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?

Solution :

(i) In a deck of 52 cards, 13 cards are spades and 4 cards are aces.

∴ P(E) = P(the card drawn is a spade) =13/52 = 1/4

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

In the deck of cards, only 1 card is an ace of spades.

P(EF) = P(the card drawn is spade and an ace) = 1/52

P(E) × P(F) =1/4 .1/12 = 1/52 = P(EF)

⇒ P(E) × P(F) = P(EF)

Therefore, the events E and F are independent.

(ii) In a deck of 52 cards, 26 cards are black and 4 cards are kings.

∴ P(E) = P(the card drawn is black) = 26/52 = 1/2

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

In the pack of 52 cards, 2 cards are black as well as kings.

∴ P (EF) = P(the card drawn is a black king) =2/52 = 1/26

P(E) × P(F) = 1/2.1/13 = 1/26 =P(EF)

Therefore, the given events E and F are independent.

(iii) In a deck of 52 cards, 4 cards are kings, 4 cards are queens, and 4 cards are jacks.

∴ P(E) = P(the card drawn is a king or a queen) = 8/52 = 2/13

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

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

∴ P(EF) = P(the card drawn is a king or a queen, or queen or a jack)

=4/52 = 1/13

P(E) × P(F) =-2/13.2/13 = 4/169 ≠ 1/13

⇒ P(E) .P(F) ≠ P(EF)

Therefore, the given events E and F are not independent.

Question16. 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 newspapers, find the probability she reads Hindi newspaper.

Solution :
Let H denote the students who read Hindi newspaper and E denote the students who read English newspaper.

It is given that,

(i) Probability that a student reads Hindi or English newspaper is

(ii) Probability that a randomly chosen student reads English newspaper, if she reads Hindi news paper, is given by P (E|H).

(iii) Probability that a randomly chosen student reads Hindi newspaper, if she reads English newspaper, is given by P (H|E).

Choose the correct answer in the following:

Question17. 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

Solution :

When two dice are rolled, the number of outcomes is 36.

The only even prime number is 2.

Let E be the event of getting an even prime number on each die.

∴ E = {(2, 2)}

⇒ P(E) = 1/36

Hence option (D) is correct.

Question18. Two events A and B are said to 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

Solution :
P (A’ and B’)

= [1 – P (A)] . [1 – P (B)]

= P (A’). P (B’)

Hence, option (B) is correct

### Solve The Following Questions.

Question1. 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?

Solution :

The urn contains 5 red and 5 black balls.

Let a red ball be drawn in the first attempt.

∴ P (drawing a red ball)  = 5/10 = 1/2

If two red balls are added to the urn, then the urn contains 7 red and 5 black balls.

P (drawing a red ball) = 7/12

Let a black ball be drawn in the first attempt.

∴ P (drawing a black ball in the first attempt)  = 5/10 = 1/2

If two black balls are added to the urn, then the urn contains 5 red and 7 black balls.

P (drawing a red ball) = 5/12

Therefore, probability of drawing second ball as red is Question2. 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.

Solution :

Let E1 and E2 be the events of selecting first bag and second bag respectively.

Let A be the event of getting a red ball.

The probability of drawing a ball from the first bag, given that it is red, is given by P (E2|A).

By using Bayes’ theorem, we obtain

Question3. 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 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 hostler?

Solution :

Let E1 and E2 be the events that the student is a hostler and a day scholar respectively and A be the event that the chosen student gets grade A.

The probability that a randomly chosen student is a hostler, given that he has an A grade, is given by P(E1|A).

By using Bayes’ theorem, we obtain

Question4. 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 the answer, will be correct with probability 1/4. What is the probability that a student knows the answer given that he answered it correctly?

Solution :

Let E1 and E2 be the respective events that the student knows the answer and he guesses the answer.

Let A be the event that the answer is correct.

The probability that the student answered correctly, given that he knows the answer, is 1.

∴ P (A|E1) = 1

Probability that the student answered correctly, given that he guessed, is 1/4.

The probability that the student knows the answer, given that he answered it correctly, is given by P(E1|A).

By using Bayes’ theorem, we obtain

Question5. 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?

Solution :
Let E1 = The person selected is suffering from certain disease, E2 = The person selected is not suffering from certain disease and A = The doctor diagnoses correctly

Since E1 and E2 are events complimentary to each other,

∴ P (E1) + P (E2) = 1

⇒ P (E2) = 1 − P (E1) = 1 − 0.001 = 0.999

Let A be the event that the blood test result is positive.

Probability that a person has a disease, given that his test result is positive, is given by

P (E1|A).

By using Bayes’ theorem, we obtain

Question6. There are three coins. One is a two headed coin, 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 head, what is the probability that it was the two headed coin?

Solution :
Let E1 = a two headed coin, E2 = a biased coin, E3 = an unbiased coin and A = A head is shown

Let A be the event that the coin shows heads.

Probability of heads coming up, given that it is a biased coin= 75%

Since the third coin is unbiased, the probability that it shows heads is always 1/2..

The probability that the coin is two-headed, given that it shows heads, is given by

P (E1|A).

By using Bayes’ theorem, we obtain

Question7. An insurance company insured 2000 scooter driver, 4000 car drivers and 6000 truck drivers. The probability of 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?

Solution :
Let E1 = Person chosen is a scooter driver, E2 = Person chosen is a car driver, E3 = Person chosen is a truck driver and A = Person meets with an accident

Let A be the event that the person meets with an accident.

There are 2000 scooter drivers, 4000 car drivers, and 6000 truck drivers.

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

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

By using Bayes’ theorem, we obtain

Question8. 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?

Solution :

Let E1 and E2 be the respective events of items produced by machines A and B. Let X be the event that the produced item was found to be defective.

∴ Probability of items produced by machine A, P (E1) = 60% = 3/5

Probability of items produced by machine B, P (E2) = 40% = 2/5

Probability that machine A produced defective items, P (X|E1)  = 2% = 2/100

Probability that machine B produced defective items, P (X|E2)  = 1% = 1/100

The probability that the randomly selected item was from machine B, given that it is defective, is given by P (E2|X).

By using Bayes’ theorem, we obtain

Question9. 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 if 0.3, if the second group wins. Find the probability that the new product introduced was by the second group.

Solution :

Let E1 and E2 be the respective events that the first group and the second group win the competition. Let A be the event of introducing a new product.

P (E1) = Probability that the first group wins the competition = 0.6

P (E2) = Probability that the second group wins the competition = 0.4

P (A|E1) = Probability of introducing a new product if the first group wins = 0.7

P (A|E2) = Probability of introducing a new product if the second group wins = 0.3

The probability that the new product is introduced by the second group is given by

P (E2|A).

By using Bayes’ theorem, we obtain

Question10. 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 noted whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 and 4 with the die?

Solution :

Let E1 be the event that the outcome on the die is 5 or 6 and E2 be the event that the outcome on the die is 1, 2, 3, or 4.

Let A be the event of getting exactly one head.

P (A|E1) = Probability of getting exactly one head by tossing the coin three times if she gets 5 or 6 = 3/8

P (A|E2) = Probability of getting exactly one head in a single throw of coin if she gets 1, 2, 3, or 4 = 1/2

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

By using Bayes’ theorem, we obtain

Question11. 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 on the job for 30% of the time and C on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?

Solution :
Let E= the item is manufactured by the operator A, E2 = the item is manufactured by the operator B, E= the item is manufactured by the operator C and A = the item is defective

Let X be the event of producing defective items.

The probability that the defective item was produced by A is given by P (E1|A).

By using Bayes’ theorem, we obtain

Question12. 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.

Solution :
Let E1 = the missing card is a diamond, E2 = the missing card is a spade, E= the missing card is a club, E4 = the missing card is a heart and A = drawing of two heart cards from the remaining cards.

Let E1 and Ebe the respective events of choosing a diamond card and a card which is not diamond.

Let A denote the lost card.

Out of 52 cards, 13 cards are diamond and 39 cards are not diamond.

When one diamond card is lost, there are 12 diamond cards out of 51 cards.

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

Similarly, 2 diamond cards can be drawn out of 51 cards in ways. The probability of getting two cards, when one diamond card is lost, is given by P (A|E1).

When the lost card is not a diamond, there are 13 diamond cards out of 51 cards.

Two cards can be drawn out of 13 diamond cards in ways whereas 2 cards can be drawn out of 51 cards in ways.

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

The probability that the lost card is diamond is given by P (E1|A).

By using Bayes’ theorem, we obtain

= 11/2/25 = 11/50

Question13. Probability that A speaks truth is 4/5 A coin is tossed. A report that a head appears. The probability that actually there was head is:

(A) 4/5

(B) 1/2

(C) 1/5

(D) 2/5

Solution :
Let A be the event that the man reports that head occurs in tossing a coin and let E1 be the event that head occurs and E2 be the event head does not occur.

E1: A speaks truth

E2: A speaks false

Let X be the event that a head appears.

P(E1)=45

Therefore,

P(E2)=1-P(E1)=1-45=15

If a coin is tossed, then it may result in either head (H) or tail (T).

The probability of getting a head is 1/2 whether A speaks truth or not.

The probability that there is actually a head is given by P (E1 | X)

Hence, option (A) is correct.

Question14. If A and B are two events such that A ⊂ B and P (B) ≠ 0, then which of the following is correct:

Solution :

If A ⊂ B, then A ∩ B = A

⇒ P (A ∩ B) = P (A)

Also, P (A) < P (B)

Hence, option (C) is correct.

### Solve The Following Questions.

Question1. State which of the following are not the probability distributions of a random variable. Give reasons for your answer.

(i)

(ii)

(iii)

(iv)

Solution :

It is known that the sum of all the probabilities in a probability distribution is one.

(i) Sum of the probabilities = 0.4 + 0.4 + 0.2 = 1

Therefore, the given table is a probability distribution of random variables.

(ii) It can be seen that for X = 3, P (X) = −0.1

It is known that probability of any observation is not negative. Therefore, the given table is not a probability distribution of random variables.

(iii) Sum of the probabilities = 0.6 + 0.1 + 0.2 = 0.9 ≠ 1

Therefore, the given table is not a probability distribution of random variables.

(iv) Sum of the probabilities = 0.3 + 0.2 + 0.4 + 0.1 + 0.05 = 1.05 ≠ 1

Therefore, the given table is not a probability distribution of random variables.

Question2. An urn contains 5 red and 2 black balls. Two balls are randomly selected. Let X represents the number of black balls. What are the possible values of X? Is X a random variable?

Solution :
There two balls may be selected as BR, RB, BR, BB, where R represents red ball and B represents black ball.

Variable X has the value 0, 1, 2, i.e., there may be no black ball, may be one black ball or both the balls are black.

Question3. Let X represents 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?

Solution :

A coin is tossed six times and X represents the difference between the number of heads and the number of tails.

∴ X (6 H, 0T) = |6 – 0| = 6

X (5 H, 1 T)  = |5 – 1| = 4

X (4 H, 2 T)  = |4 – 2| = 2

X (3 H, 3 T)  = 3 – 3| = 0

X (2 H, 4 T)  = |2 – 4| = 2

X (1 H, 5 T)  = |1 – 5| = 4

X (0H, 6 T) = | 0 – 6| = 6

Thus, the possible values of X are 6, 4, 2, and 0.

Question4. 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.

Solution :

(i) When one coin is tossed twice, the sample space is

{HH, HT, TH, TT}

Let X represent the number of heads.

∴ X (HH) = 2, X (HT) = 1, X (TH) = 1, X (TT) = 0

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

It is known that,

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

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

P (X = 1) = P (HT) + P (TH) = 14 + 1/4 = 1/2

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

Thus, the required probability distribution is as follows.

(ii) When three coins are tossed simultaneously, the sample space is {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}

Let X represent the number of tails.

It can be seen that X can take the value of 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

Thus, the probability distribution is as follows.

(iii) When a coin is tossed four times, the sample space is

(iii) When a coin is tossed four times, the sample space is

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

Let X be the random variable, which represents the number of heads.

It can be seen that X can take the value of 0, 1, 2, 3, or 4.

P (X = 0) = P (TTTT) = 1/16

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

= 1/16 + 1/16 + 1/16 + 1/16 = 4/16 = 1/4

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

+ P (THTH)

= 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 = 6/16 = 3/8

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

= 1/16 + 1/16 + 1/16 + 1/16 = 4/16 = 1/4

P (X = 4) = P (HHHH) = 1/16

Thus, the probability distribution is as follows.

Question5. Find the probability distribution of the number of success 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.

Solution :

When a die is tossed two times, we obtain (6 × 6) = 36 number of observations.

Let X be the random variable, which represents the number of successes.

(i) Here, success refers to the number greater than 4.

P (X = 0) = P (number less than or equal to 4 on both the tosses) = 4/6 X 4/6 = 4/9

P (X = 1) = P (number less than or equal to 4 on first toss and greater than 4 on second toss) + P (number greater than 4 on first toss and less than or equal to 4 on second toss)

= 4/6 X 2/6 + 4/6 X 2/6 = 4/9

P (X = 2) = P (number greater than 4 on both the tosses)

= 2/6 X 2/6 = 1/9

Thus, the probability distribution is as follows.

(ii) Here, success means six appears on at least one die.

P (Y = 0 ) = P (six appears on  none of the dice) = 5/6 x  5/6 = 25/36

P (Y = 1) = P (six appears on at least one of the dice) = 1/6 x 5/6 + 5/6 x 1/6 + 1/6 x 1/6 = 11/36

Thus, the required probability distribution is as follows.

Question6. 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.

Solution :

It is given that out of 30 bulbs, 6 are defective.

⇒ Number of non-defective bulbs = 30 − 6 = 24

4 bulbs are drawn from the lot with replacement.

Let X be the random variable that denotes the number of defective bulbs in the selected bulbs.

Therefore, the required probability distribution is as follows.

Question7. A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.

Solution :

Let the probability of getting a tail in the biased coin be x.

∴ P (T) = x

⇒ P (H) = 3x

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

When the coin is tossed twice, the sample space is {HH, TT, HT, TH}.

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

∴ P (X = 0) = P (no tail) = P (H) × P (H) = 3/4 x 3/4 = 9/16

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

= 3/4 . 1/4 + 1/4 . 3/4

= 3/16 + 3/16= 3/8

P (X = 2) = P (two tails) = P (TT) = 1/4 x 1/4 = 1/16

Therefore, the required probability distribution is as follows.

Question8. A random variable X has the following probability distribution:

Determine:

(i) k

(ii) P (X < 3)

(iii) P (X > 6)

(iv) P (0 < X < 3)

Solution :
(i) Since, the sum of all the probabilities of a distribution is 1.

Question9. The random variable X has a probability distribution P(X)  of the following form, where k is some number:

(a) Determine the value of k.

(b) Find P(X < 2), P(X ≥ 2), P(X ≥ 2).

Solution :

(a) It is known that the sum of probabilities of a probability distribution of random variables is one.

∴ k + 2k + 3k + 0 = 1

⇒ 6k = 1

⇒ k =1/6

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

= k + 2k

= 3k

=3/6 = 1/2

10. Find the mean number of heads in three tosses of fair coin.

Solution :

Let X denote the success of getting heads.

Therefore, the sample space is

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

It can be seen that X can take the value of 0, 1, 2, or 3

Therefore, the required probability distribution is as follows.

Question11. Two dice are thrown simultaneously. If X denotes the number of sixes, find expectation of X.

Solution :
Two dice thrown simultaneously is the same the die thrown 2 times.

Here, X represents the number of sixes obtained when two dice are thrown simultaneously. Therefore, X can take the value of 0, 1, or 2.

∴ P (X = 0) = P (not getting six on any of the dice) = 25/36

P (X = 1) = P (six on first die and no six on second die) + P (no six on first die and six on second die)

=2(1/6 x 5/6) = 10/36

P (X = 2) = P (six on both the dice) =1/36

Therefore, the required probability distribution is as follows.

Then, expectation of X = E(X) =∑ X iP(Xi)

= 0 x 25/36 + 1 x 10/36 + 2 x 1/36

= 1/3

Question12. Two numbers are selected at random (without replacement), from the first six positive integers. Let X denotes the larger of two numbers obtained. Find E (X).

Solution :

The two positive integers can be selected from the first six positive integers without replacement in 6 × 5 = 30 ways

X represents the larger of the two numbers obtained. Therefore, X can take the value of 2, 3, 4, 5, or 6.

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

P(X = 2) = 2/30 = 1/15

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

P(X = 3) = 4/30 = 2/15

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

P(X = 4) = 6/30 = 1/5

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

P(X = 5) = 8/30 = 4/15

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

P(X = 6) = 10/30 = 1/3

Therefore, the required probability distribution is as follows.

Question13. Let X denotes the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of X.

Solution :

When two fair dice are rolled, 6 × 6 = 36 observations are obtained.

P(X = 2) = P(1, 1) = 1/36

P(X = 3) = P (1, 2) + P(2, 1) = 2/36 = 1/18

P(X = 4) = P(1, 3) + P(2, 2) + P(3, 1) = 3/36 = 1/12

P(X = 5) = P(1, 4) + P(2, 3) + P(3, 2) + P(4, 1) = 4/36 = 1/9

P(X = 6) = P(1, 5) + P (2, 4) + P(3, 3) + P(4, 2) + P(5, 1) = 5/36

P(X = 7) = P(1, 6) + P(2, 5) + P(3, 4) + P(4, 3) + P(5, 2) + P(6, 1) = 6/36 = 1/6

P(X = 8) = P(2, 6) + P(3, 5) + P(4, 4) + P(5, 3) + P(6, 2) = 5/36

P(X = 9) = P(3, 6) + P(4, 5) + P(5, 4) + P(6, 3) = 4/36 = 1/9

P(X = 10) = P(4, 6) + P(5, 5) + P(6, 4) = 3/36 = 1/12

P(X = 11) = P(5, 6) + P(6, 5) = 2/36 = 1/18

P(X = 12) = P(6, 6) = 1 x 36

Therefore, the required probability distribution is as follows.

Question14. A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable X? Find mean, variance and standard deviation of X.

Solution :

There are 15 students in the class. Each student has the same chance to be chosen. Therefore, the probability of each student to be selected is 1/15.

The given information can be compiled in the frequency table as follows.

P(X = 14) =2/15, P(X = 15) =1/15, P(X = 16) =2/15, P(X = 16) =3/15,

P(X = 18) =115, P(X = 19) =2/15, P(X = 20) =3/15, P(X = 21) =1/15

Therefore, the probability distribution of random variable X is as follows.

Then, mean of X = E(X)

Question15. In a meeting 70% of the members favour a certain proposal, 30% being opposed. A member is selected at random and we let X = 0 if the opposed and X = 1 if he is in favour. Find E (X) and Var (X).

Solution :

It is given that P(X = 0) = 30% = 30/100 = 0.3

P(X = 1) = 70% = 70/100 = 0.7

Therefore, the probability distribution is as follows.

= 0.7 − (0.7)2

= 0.7 − 0.49

= 0.21

Choose the correct answer in each of the following:

Question16. The mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is:

(A) 1

(B) 2

(C) 5

(D) 8/3

Solution :

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

The total number of observations is six.

Therefore, the probability distribution is as follows.

Therefore, option (B) is correct.

Question 17. Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. What is the value of E(X)?

(A) 37/221

(B) 5/13

(C) 1/13

(D) 2/13

Solution :

Let X denote the number of aces obtained. Therefore, X can take any of the values of 0, 1, or 2.

In a deck of 52 cards, 4 cards are aces. Therefore, there are 48 non-ace cards.

Therefore, option (D) is correct.

### Solve The Following Questions.

Question1.  A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of:

(i)  5 successes?

(ii)  at least 5 successes?

(iii)  at most 5 successes?

Solution :

The repeated tosses of a die are Bernoulli trials. Let X denote the number of successes of getting odd numbers in an experiment of 6 trials.

Probability of getting an odd number in a single throw of a die is, p – 3/6 = 1/2

q = 1 – p = 1/2

X has a binomial distribution.

Question2.  A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of two successes.

Solution :

The repeated tosses of a pair of dice are Bernoulli trials. Let X denote the number of times of getting doublets in an experiment of throwing two dice simultaneously four times.

Probability of getting doublets in a single throw of the pair of dice is

Question3.  There are 5% defective items in a large bulk of items. What is the probability that a sample of 10 items will include not more than one defective item?

Solution :

Let X denote the number of defective items in a sample of 10 items drawn successively. Since the drawing is done with replacement, the trials are Bernoulli trials.

Question4.  Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. What is the probability that:

(i)  all the five cards are spade?

(ii)  only 3 cards are spades?

Solution :

Let X represent the number of spade cards among the five cards drawn. Since the drawing of card is with replacement, the trials are Bernoulli trials.

In a well shuffled deck of 52 cards, there are 13 spade cards.

Question5.  The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs.

Solution :

Let X represent the number of bulbs that will fuse after 150 days of use in an experiment of 5 trials. The trials are Bernoulli trials.

It is given that, p = 0.05

Question6.  A bag consists of 10 balls each marked with one of the digits 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0?

Solution :

Let X denote the number of balls marked with the digit 0 among the 4 balls drawn.

Since the balls are drawn with replacement, the trials are Bernoulli trials.

X has a binomial distribution with n = 4 and p = 1/10

Question7.  In an examination, 20 questions of true-false are asked. Suppose a student tosses a fair coin to determine his answer to each question. If the coin falls heads, he answers ‘true’, if it falls tails, he answers ‘false’. Find the probability that he answers at least 12 questions correctly.

Solution :

Let X represent the number of correctly answered questions out of 20 questions.

The repeated tosses of a coin are Bernoulli trails. Since “head” on a coin represents the true answer and “tail” represents the false answer, the correctly answered questions are Bernoulli trials.

Question8.  Suppose X has a binomial distribution B(6,1/2) Show that X = 3 is the most likely outcome.

(Hint: P(X = 3) is the maximum among all P (xi), xi = 0, 1, 2, 3, 4, 5, 6)

Solution :

X is the random variable whose binomial distribution is B(6,1/2).

Therefore, n = 6 and p = 1/2

Therefore, P (X = 3) is maximum.

Question9.  On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?

Solution :

The repeated guessing of correct answers from multiple choice questions are Bernoulli trials. Let X represent the number of correct answers by guessing in the set of 5 multiple choice questions.

Probability of getting a correct answer is, p = 1/3

Question10.  A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is 1/100  What is the probability that he will win a prize:

(a) at least once

(b) exactly once

(c) at least twice?

Solution :

Let X represent the number of winning prizes in 50 lotteries. The trials are Bernoulli trials.

Clearly, X has a binomial distribution with n = 50 and p = 1/100

Question11.  Find the probability of getting 5 exactly twice in 7 throws of a die.

Solution :

The repeated tossing of a die are Bernoulli trials. Let X represent the number of times of getting 5 in 7 throws of the die.

Probability of getting 5 in a single throw of the die, = 1/6

Question12.  Find the probability of throwing at most 2 sixes in 6 throws of a single die.

Solution :

The repeated tossing of the die are Bernoulli trials. Let X represent the number of times of getting sixes in 6 throws of the die.

Probability of getting six in a single throw of die, p = 1/6

Let A represents the favourable event i.e., 6

Question13.  It is known that 10% of certain articles manufactured are defective. What is the probability that in a random sample of 12 such articles 9 are defective?

Solution :

The repeated selections of articles in a random sample space are Bernoulli trails. Let X denote the number of times of selecting defective articles in a random sample space of 12 articles.

Clearly, X has a binomial distribution with n = 12 and = 10% = 10/100 = 1/10

In each of the following, choose the correct answer:

Question14.  Binomial distribution is given this name because:

(A)  This distribution was evolved by James binomial.

(B)  Each trial has only two outcomes. Namely success and failure.

(C)  Its probability function is obtained by general of binomial expansion.

(D)  It is obtained by combining two distributions.

Solution: option (C) is correct.

Question15.  In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is:

(D) None of these

Solution :

The repeated selection of students who are swimmers are Bernoulli trials. Let X denote the number of students, out of 5 students, who are swimmers.

Probability of students who are not swimmers, q = 1/5

Therefore, option (A) is correct.

### Solve The Following Questions.

Question1.A and B are two events such that P (A) ≠ 0. Find P (B/A) if:

(i)A is a subset of B

(ii) A ∩ B = Φ

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

To find: P(A)

(i) A is a subset of B

(ii) A ∩ B = Φ

Question2.A couple has two children.

(i)Find the probability that both children are males I it is known that at least one of the children is male.

(ii)Find the probability that both children are females if it is known that the elder child is a female.

Solution :

If a couple has two children, then the sample space is

S = {(bb), (bg), (gb), (gg)}

(i) Let E and F respectively denote the events that both children are males and at least one of the children is a male.

(ii) Let A and B respectively denote the events that both children are females and the elder child is a female.

Question3.Suppose that 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.

Solution :

It is given that 5% of men and 0.25% of women have grey hair.

Therefore, percentage of people with grey hair = (5 + 0.25) % = 5.25%

∴ Probability that the selected haired person is a male = 5/5.25 = 20/21

Question4.Suppose that 90% of people are right-handed. What is the probability that at most of 6 of a random sample of 10 people are right-handed?

Solution :

A person can be either right-handed or left-handed.

It is given that 90% of the people are right-handed.

Using binomial distribution, the probability that more than 6 people are right-handed is given by,

Therefore, the probability that at most 6 people are right-handed

= 1 − P (more than 6 are right-handed)

Question5.An urn contains 25 balls of which 10 balls bear a mark ‘X’ and the remaining 15 bear a mark ‘Y’. A ball is drawn at random from the urn, its mark noted down and it is replaced. If 6 balls are drawn in this way, find the probability that:

(i)all will bear ‘X’ mark.

(ii)not more than 2 will bear ‘Y’ mark.

(iii)at least one ball will bear ‘Y’ mark.

(iv)the number of balls with ‘X’ mark and ‘Y’ mark will be equal.

Solution :

Total number of balls in the urn = 25

Balls bearing mark ‘X’ = 10

Balls bearing mark ‘Y’ = 15

p = P (ball bearing mark ‘X’) =10/25 = 2/5

q = P (ball bearing mark ‘Y’) =15/25 = 3/5

Six balls are drawn with replacement. Therefore, the number of trials are Bernoulli trials.

Let Z be the random variable that represents the number of balls with ‘Y’ mark on them in the trials.

Clearly, Z has a binomial distribution with n = 6 and p =2/5.

Question6.In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is 5/6  What is the probability that he will knock down fewer than 2 hurdles?

Solution :

Let p and q respectively be the probabilities that the player will clear and knock down the hurdle.

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

Therefore, by binomial distribution, we obtain

Question7.A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die.

Solution :

The probability of getting a six in a throw of die is 1/6 and not getting a six is 5/6.

Let p = 1/6 and q = 5/6

The probability that the 2 sixes come in the first five throws of the die is

∴ Probability that third six comes in the sixth throw =

Question8.If a leap year is selected at random, what is the change that it will contain 52 Tuesday?

Solution :
A leap year has 366 days which means 52 complete weeks and 2 days. If any one of these two days in a Tuesday then the year will have 53 Tuesdays.

Number of total days in a week = 7

Number of favourable days = 2

Therefore, P (the year will have 53 Tuesday) = 2/7

Question9.An experiment succeeds twice as often as it fails. Find the probability that in the next six trails, there will be at least 4 successes.

Solution :

The probability of success is twice the probability of failure.

Let the probability of failure be x.

∴ Probability of success = 2x

Let X be the random variable that represents the number of successes in six trials.

By binomial distribution, we obtain

Question10.How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?

Solution :

Let the man toss the coin n times. The n tosses are n Bernoulli trials.

Probability (p) of getting a head at the toss of a coin is 1/2.

∴ p = 1/2 ⇒ q = 1/2

It is given that,

P (getting at least one head) > 90/100

P (x ≥ 1) > 0.9

⇒ 1 − P (x = 0) > 0.9

The minimum value of n that satisfies the given inequality is 4.

Thus, the man should toss the coin 4 or more than 4 times.

Question11.In a game a man wins a rupee for a six and looses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amounts he wins/looses.

Solution :
When a die is thrown, then probability of getting a six = 16

then, probability of not getting a six = 1 – 16 = 56

If the man gets a six in the first throw, then

probability of getting a six = 16

If he does not get a six in first throw, but gets a six in second throw, then

probability of getting a six in the second throw = 56×16 = 536

If he does not get a six in the first two throws, but gets in the third throw, then

probability of getting a six in the third throw = 56×56×16 = 25216

probability that he does not get a six in any of the three throws = 56×56×56 = 125216

In the first throw he gets a six, then he will receive Re 1.

If he gets a six in the second throw, then he will receive Re (1 – 1) = 0

If he gets a six in the third throw, then he will receive Rs(-1 – 1 + 1) = Rs (-1),

that means he will lose Re 1 in this case.

Expected value = 16×1 + 56×16 × 0 + 56×56×16×-1 = 11216

So, he will loose Rs 11216.

Question12.Suppose we have four boxes A, B, C and D containing coloured marbles as given below:

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B? box C?

Solution :

Let R be the event of drawing the red marble.

Let EA, EB, and EC respectively denote the events of selecting the box A, B, and C.

Total number of marbles = 40

Number of red marbles = 15

P(R) = 15/40 = 3/8

Probability of drawing the red marble from box A is given by P (EA|R).

Probability that the red marble is from box B is P (EB|R).

Probability that the red marble is from box C is P (EC|R).

Question13.Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduces the risk of heart attack by 30% and prescription of certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga.

Solution :
A patient has options to have the treatment of yoga and meditation and that of prescription of drugs.

Let these events be denoted by Eand E2 i.e.,

E1 = Treatment of yoga and meditation

E2 = Treatment of prescription of certain drugs

P (E1) = P (E2) = 1/2

Let A denotes that a person has heart attack, then P (A) = 40% = 0.40

Yoga and meditation reduces heart attack by 30.

Inspite of getting yoga and meditation treatment heart risk is 70% of 0.40

P (A|E1)  = 0.40 x 0.70 = 0.28

Also, Drug prescription reduces the heart attack rick by 25%

Even after adopting the drug prescription hear rick is 75% of 0.40

P (A|E2)  = 0.40 x 0.75 = 0.30

Question14.If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability 1/2).

Solution :
There are four entries in a determinant of 2 x 2 order. Each entry may be filled up in two ways with 0 or 1.

Number of determinants that can be formed = (2)4 = 16

The value of determinants is positive in the following cases:

Therefore, the probability that the determinant is positive = 3/16

Question15. An electronic assembly consists of two sub-systems say A and B. From previous testing procedures, the following probabilities are assumed to be known:

P (A fails) = 0.2

P (B fails alone) = 0.15

P (A and B fail) = 0.15

Evaluate the following probabilities.

Solution :

Let the event in which A fails and B fails be denoted by EA and EB.

P (EA) = 0.2

P (E∩ EB) = 0.15

P (B fails alone) = P (EB) − P (EA ∩ EB)

⇒ 0.15 = P (EB) − 0.15

⇒ P (EB) = 0.3

(ii) P (A fails alone) = P (EA) − P (E∩ EB)

= 0.2 − 0.15

= 0.05

Question16.Bag I contains 3 Red and 4 Black balls and B II contains 4 Red and % Black balls. One ball is transferred from Bag I to bag II and then a ball is drawn from bag II. The ball so drawn is found to be Red in colour. Find the probability that the transferred ball is Black.

Solution :
Let E= Ball transferred from Bag I to Bag II is red

E= Ball transferred from Bag I to Bag Ii is black

A = Ball drawn from Bag II is red in colour

P (E1) =3/7 and P (E2) = 4/7

Let A be the event that the ball drawn is red.

When a red ball is transferred from bag I to II,

P (A | E1) = 5/10 = 1/2

When a black ball is transferred from bag I to II,

P (A | E2) = 4/10 = 2/5

Choose the correct answer in each of the following:

Question17.If A and B are two events such that P (A) ≠ 0 and P(B|A) = 1, then:

(A) A ⊂ B

(B) B ⊂ A

(C) B = Φ

(D) A = Φ

Solution : Therefore, option (A) is correct.

Question18. If P (A|B) > P (A), then which of the following is correct:

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

(B) P (A ∩ B) < P (A).P (B)

(C) P (B|A) > P (B)

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

Solution : Therefore, option (C) is correct.

Question19.If A and B are any two events such that P (A) + P (B) – P (A and B) = P (A), then

(A) P (B|A) = 1

(B) P (A|B) = 1

(C) P (B|A) = 0

(D)P (A|B) = 0

Solution : Therefore, option (B) is correct.

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