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

53. 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.

Answer:

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

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

⇒ P (T) = x

And P (H) = 3x

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

⇒ x + 3x = 1

⇒ 4x = 1

⇒ x = 1/4

Hence, P (T) = 1/4 and P (H) = 3/4

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

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

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

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

P(X = 1) = P (one tail) = P (HT) × P (TH) = 3/4. 1/4 × 1/4. 3/4 = 3/8

P(X = 2) = P (two tail) = P (T) × P (T) = 1/4 × 1/4 = 1/16

Hence, the required probability distribution is,

X	0	1	2
P (X)	9/16	3/8	1/16

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

X	0	1	2	3	4	5	6	7
P (X)	0	k	2k	2k	3k	k2	2k2	7k2 + k

Determine

(i) k

(ii) P (X < 3)

(iii) P (X > 6)

(iv) P (0 < X < 3)

Answer:

Given a random variable X with its probability distribution.

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

Hence the sum of probabilities of given table:

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

⇒ 10K² + 9k = 1

⇒ 10K² + 9k – 1 = 0

⇒ (10K-1) (k + 1) = 0

k = -1, 1/10

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

So k = 1/10

(ii) Now we have to find P(X < 3)

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

= 0 + k + 2k

= 3k

P (X < 3) = 3 × 1/10 = 3/10

(iii) Now we have find P(X > 6)

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

= 7K² + k

= 7 × (1/10)² + 1/10

= 7/100 + 1/10

P (X > 6) = 17/100

(iv) Consider P (0 < X < 3)

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

= k + 2k

= 3k

P (0 < X < 3) = 3 × 1/10 = 3/10

55. 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).

Answer:

Given: A random variable X with its probability distribution.

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

Hence the sum of probabilities of given table:

⇒ k + 2k + 3k + 0 = 1

⇒ 6k = 1

k = 1/6

(b) Now we have to find P(X < 2)

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

= k + 2k

= 3k

P (X < 2) = 3 × 1/6 = 1/2

Consider P (X ≤ 2)

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

= k + 2k + 3k

= 6k

P (X ≤ 2) = 6 × 1/6 = 1

Now we have to find P(X ≥ 2)

P(X ≥ 2) = P(X = 2) + P(X > 2)

= 3k + 0

= 3k

P (X ≥ 2) = 3 × 1/6 = 1/2

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

Answer:

Given a coin is tossed three times.

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

X represents the number of heads.

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

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

P (X = 0) = P (TTT) = 1/8

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

P (X = 2) = P (THH) + P (HTH) + P (HHT) = 1/8 + 1/8 + 1/8 = 3/8

P (X = 3) = P (HHH) = 1/8

Hence, the required probability distribution is,

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

 Therefore mean μ is:

Now by substituting the values we get

Answer 2

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

Answer:

Given a die is thrown two times.

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

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

Now

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

P (X = 2) = P (six does appear on both of die) = 1/6 × 1/6 = 1/36

Hence, the required probability distribution is,

X	0	1	2
P (X)	25/36	5/18	1/36

 Therefore Expectation of X E (X):

Now by substituting the values we get

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

Answer:

Given first six positive integers.

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

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

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

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

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

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

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

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

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

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

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

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

Hence, the required probability distribution is,

X	2	3	4	5	6
P (X)	1/15	2/15	1/5	4/15	1/3

 Therefore E(X) is:

Now by substituting the values we get

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

Answer:

Given two fair dice are rolled

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

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

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

P (X) = 1/36

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

P (X) = 2/36 = 1/18

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

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

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

P (X) = 4/39 = 1/9

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

P (X) = 5/36

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

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

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

P (X) = 5/36

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

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

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

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

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

P (X) = 2/36 = 1/18

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

P (X) = 1/36

Hence, the required probability distribution is,

X	2	3	4	5	6	7	8	9	10	11	12
P(X)	1/36	1/18	1/12	1/9	5/36	1/6	5/36	1/9	1/12	1/18	1/36

 Therefore E(X) is:

Now by substituting the values we get

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

= 54.833 - (7)²

= 54.833 - 49

= 5.833

And Standard deviation = \(\sqrt{Var(X)}\)

= \(\sqrt{5.833}\)

= 2.415

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

Answer:

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

P(X = 0) = 30% = 30/100 = 0.3

P(X = 1) = 70% = 70/100 = 0.7

Hence, the required probability distribution is,

X	0	1
P (X)	0.3	0.7

 Therefore E(X) is:

= 0 × 0.3 + 1 × 0.7

⇒ E(X) = 0.7

And E(X²) is:

= (0)² × 0.3 + (1)² × 0.7

⇒ E(X²) = 0.7

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

= 0.7 – (0.7)²

= 0.7 – 0.49 = 0.21

Answer 3

61. 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

Answer:

Correct option is B. 2

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

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

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

The total numbers is six.

Now

P(X = 1) = 3/6 = ½

P(X = 2) = 1/3

P(X = 5) = 1/6

Hence, the required probability distribution is,

X	1	2	5
P(X)	1/2	1/3	1/6

 Therefore Expectation of X E(X):

Now by substituting the values we get

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

A. 37/221

B. 5/13

C. 1/13

D. 2/13

Answer:

Correct option is D. 2/13

Given a deck of cards.

X be the number of aces obtained.

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

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

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

\(= \frac {C_0 \times 48C_2}{52C_2}\)

\(= \frac{1128}{1326}\)

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

\(= \frac {4C_1 \times 48C_1}{52C_2}\)

\(= \frac{192}{1326}\)

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

\(=\frac{6}{1326}\)

Hence, the required probability distribution is,

X	0	1	2
P(X)	1128/1326	192/1326	6/1326

 Therefore Expectation of X E(X):

63. 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?

Answer:

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

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

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

\(= \frac{\text{number of odd numbers on a dice}}{\text{total number of numbers on a dice }}\)

\(= \frac36 = \frac12\)

Thus, q = 1 – p = 1/2

Now, here x has a binomial distribution.

Thus, P(X = x) = ⁿC_x q^n-x p^x, where x = 0, 1, 2 …n

= ⁶C_x (1/2)^6-x (1/2)^x

= ⁶C_x (1/2)⁶

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

= ⁶C₅ (1/2)⁶

= 6 ×1/64

= 3/32

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

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

= ⁶C₅ (1/2)⁶ + ⁶C₅ (1/2)⁶

= 6 ×1/64 + 6 ×1/64

= 6/64 + 1/64

= 7/64

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

We can also write it as: 1 – P(X > 5)

= 1 – P(X = 6)

= 1 – ⁶C₆ (1/2)⁶

= 1 – 1/64

= 63/64

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

Answer:

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

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

Probability of getting doublets in a single throw of a pair of dice (p)

\(= \frac{\text{number of doublets possible in a pair of dice}}{\text{total number of possible pairs when two dice thrown}}\)

\(= \frac6{36} = \frac16\)

Thus, q = 1 – p = 1 – 1/6 = 5/6

Now, here x has a binomial distribution, where n = 4, p = 1/6, q = 5/6

Thus, P(X = x) = ⁿC_x q^n-x p^x, where x = 0, 1, 2, … n

= ⁴C_x (5/6)^4-x (1/6)^x

= ⁴C_x (5^4-x/6⁶)

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

= ⁴C₂ (5^4-2/6⁴)

= 6 (5²/6⁴)

= 6 × (25/1296)

= 25/216

65. 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?

Answer:

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

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

Now, probability of getting a defective item, p = 5/100 = 1/20

Thus, q = 1 – 1/20 = 19/20

∴ We can say that x has a binomial distribution, where n = 10 and p = 1/20

Thus, P(X = x) = ⁿC_x q^n-x p^x, where x = 0, 1, 2 …n

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

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

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

Answer 4

66. 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 spades?

(ii) Only 3 cards are spades?

(iii) None is a spade?

Answer:

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

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

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

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

= 13/52 = ¼

q = 1 – ¼ = 3/4

Thus, x has a binomial distribution with n = 5 and p = ¼

Thus, P(X = x) = ⁿC_x q^n-x p^x , where x = 0, 1, 2, …n

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

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

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

67. 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

(i) none

(ii) not more than one

(iii) more than one

(iv) at least one will fuse after 150 days of use.

Answer:

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

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

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

Thus, q = 1 – p = 1 – 0.05 = 0.95

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

Thus, P(X = x) = ⁿC_x q^n-x p^x, where x = 0, 1, 2… n

= ⁵C_x (0.95)^5-x(0.05)^x

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

= ⁵C₀ (0.95)^5-0(0.05)⁰

= 1× 0.95⁵

= (0.95)⁵

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

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

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

= 1× 0.95⁵ + 5 × (0.95)⁴ × 0.05

= (0.95)⁴ (0.95 +0.25)

= (0.95)⁴ × 1.2

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

= 1 – P(X ≤ 1)

= 1 – [(0.95)⁴ × 1.2]

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

= 1 – P(X < 1)

= 1 – P(X = 0)

= 1 – (0.95)⁵

68. 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?

Answer:

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

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

Probability of a ball drawn from the bag to be marked as digit 0 = 1/10

It can be clearly observed that X has a binomial distribution with n = 4 and p = 1/10

Thus, q = 1 – p = 1 – 1/10 = 9/10

Thus, P(X = x) = ⁿC_x q^n-x p^x, where x = 0, 1, 2, …n

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

69. In an examination, 20 questions of true-false type 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.

Answer:

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

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

Thus, p = ½ and q = 1 – p = 1 – ½ = ½

Here, it can be clearly observed that x has binomial distribution, where n = 20 and p = ½

Thus, P(X = x) = ⁿC_x q^n-x p^x, where x = 0, 1, 2 … n

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

70. Suppose X has a binomial distribution B (6, ½)  . 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)

Answer:

Given X is any random variable whose binomial distribution is B (6, 1/2)

Thus, n = 6 and p = ½

q = 1 – p = 1 – ½ = ½

Thus, P(X = x) = ⁿC_x q^n-x p^x, where x = 0, 1, 2 …n

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

∴⁶c_x = ⁶c₆ = 1

⁶c₁ = ⁶c₅ = 6

⁶c₂ = ⁶c₄ = 15

⁶c₃ = 20

Hence we can clearly see that ⁶c₃ is maximum.

∴ for x = 3, P(X = x) is maximum.

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

71. 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?

Answer:

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

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

Now, probability of getting a correct answer, p = 1/3

Thus, q = 1 – p = 1 – 1/3 = 2/3

Clearly, we have X is a binomial distribution where n = 5 and P = 1/3

Hence, probability of guessing more than 4 correct answer = P(X ≥ 4)

Answer 5

72. 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?

Answer:

(a) Let X represents the number of prizes winning in 50 lotteries and the trials are Bernoulli trials

Here clearly, we have X is a binomial distribution where n = 50 and p = 1/100

Thus, q = 1 – p = 1 – 1/100 = 99/100

Hence, probability of winning in lottery at least once = P(X ≥ 1)

(b) Probability of winning in lottery exactly once = P(X = 1)

(c) Probability of winning in lottery at least twice = P(X ≥ 2)

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

Answer:

Let us assume X represent the number of times of getting 5 in 7 throws of the die

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

Thus, probability of getting 5 in a single throw, p = 1/6

And, q = 1 – p

= 1 – 1/6

= 5/6

Clearly, we have X has the binomial distribution where n = 7 and p = 1/6

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

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

Answer:

Let us assume X represent the number of times of getting sixes in 6 throws of a die

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

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

Clearly, we have X has the binomial distribution where n = 6 and p = 1/6

And, q = 1 – p = 1 – 1/6 = 5/6

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

75. 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?

Answer:

Let us assume X represent the number of times selecting defected articles in a random sample space of given 12 articles

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

Clearly, we have X has the binomial distribution where n = 12 and p = 10% = 1/10

And, q = 1 – p = 1 – 1/10 = 9/10

Hence, probability of selecting 9 defective articles 

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

A. 10^–1

B. (1/2)⁵

C. (9/10)⁵

D. 9/10 

Answer:

Correct option is C. (9/10)⁵

Let us assume X represent the number of times selecting defected bulbs in a random sample of given 5 bulbs

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

Clearly, we have X has the binomial distribution where n = 5 and p = 1/10

And, q = 1 – p = 1 – 1/10

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

77. The probability that a student is not a swimmer is 1/5. Then the probability that out of five students, four are swimmers is

A. ⁵C₄ 1/5 (4/5)⁴

B. (4/5)⁴ (1/5)

C. ⁵C₁ 1/5 (4/5)⁴

D. None of these

Answer:

Correct option is A. ⁵C₄ 1/5 (4/5)⁴

Let us assume X represent the number of students out of 5 who are swimmers

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

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

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

And, p = 1 – q

= 1 – 1/5

= 4/5

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

78. 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

Answer:

It is given that,

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

(i) We have,

Hence,

(ii) We have,

Answer 6

79. A couple has two children,

(i) Find the probability that both children are males, if 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.

Answer:

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

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

Assume that A denote the event of both children having male and B denote the event of having at least one of the male children

Thus, we have:

(ii) Assume that C denote the event having both children females and D denote the event of having elder child is female

∴ C = {(g, g)}

P (C) = ¼

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

P (D) = (2/4)

80. 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.

Answer:

Given that, 5% of men and 0.25% of women have grey hair

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

= 5.25 %

Hence, Probability of having a selected person male having grey hair, P = 5/25 = 20/21

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

Answer:

Given that, 90% of the people are right handed

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

p = 9/10 and q = 1 – 9/10 = 1/10

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

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

82. 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?

Answer:

Assume that p be the probability of player that will clear the hurdle while q be the probability of player that will knock down the hurdle

∴ p = 5/6 and q = 1 – 5/6 = 1/6

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

∴ By binomial distribution,

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

83. 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.

Answer:

From the given question, it is clear that

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

And, probability of not getting a six = 5/6

Let us assume, p = 1/6 and q = 5/6

Now, we have

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

Also, Probability that the six come in the sixth throw 

84. If a leap year is selected at random, what is the chance that it will contain 53 Tuesdays?

Answer:

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

Now, in 52 weeks there are total 52 Tuesdays

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

Thus, the remaining two days can be

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

∴ Total Number of cases = 7

Cases in which Tuesday can come = 2

Hence, probability (leap year having 53 Tuesdays) = 2/7

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

Answer:

Given that probability of failure = x

And, probability of success = 2x

∴ x + 2x = 1

3x = 1

X = 1/3

2x = 2/3

Assume p = 1/3 and q = 2/3

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

Hence, by binomial distribution we have:

∴ Probability of having at least 4 successes = P(X ≥ 4)

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

Answer:

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

∴ Probability of getting head at the toss of the coin = ½

Let us assume, p = 1/2 and q = 1/2

It is given in the question that,

Probability of getting at least one head > 90/100

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

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

Answer 7

87. In a game, a man wins a rupee for a six and loses 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 amount he wins / loses.

Answer:

For the situation given in the equation, we have

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

Also, probability of not getting a 6 = 5/6

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

(i) First case is that, if he gets a six on his first through then the required probability will be 1/6

∴ Amount received by him = Rs. 1

(ii) Secondly, if he gets six on his second throw then the probability = (5/6 × 1/6)

= 5/36

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

= 0

(iii) Lastly, if he does not get six in first two throws and gets six in his third throw then the probability  = 5/6 × 5/6 × 1/6

= 25/216

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

= – 1

Hence, expected value that he can win = 1/6 – 25/216

= (36 – 25)/216

= 11/216

88. Assume that the chances of a patient having a heart attack are 40%. It is also assumed that a meditation and yoga course reduce 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?

Answer:

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

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

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

It is given in the question that,

P (X) = 0.40

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

89. 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).

Answer:

rom the question, we have:

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

= 16

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

∴ Required probability = 3/16

90. Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 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.

Answer:

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

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

∴ P (A₁) = 3/7

And, P (A₂) = 4/7

Let X be the event that the drawn ball is red

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

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

91. 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

Answer:

Correct option is A. A ⊂ B

It is given in the question that,

A ad B are two events where,

92. 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)

Answer:

Correct option is C. P (B|A) > P (B)

Given that,

93. 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

Answer:

Correct option is B. P(A|B) = 1

Given that,

A and B are any two events where,