# Class 12 Maths NCERT Solutions for Chapter 13 Probability Exercise 13.5

### Probability Exercise 13.5 Solutions

**1. 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.

**2. 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

P = 6/36 = 1/6

∴ q = 1 - p = 1 - 1/6 = 5/6

Clearly, X has the binomial distribution with n = 4, P = 1/6, and q = 5/6

**3. 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.

⇒ P = 5/100 = 1/20

∴ q = 1 - 1/20 = 19/20

X has a binomial distribution with n = 10 and p = 1/20

**4. 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? **

**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.

⇒ P = 13/52 = 1/4

∴ q = 1 - 1/4 = 3/4

X has a binomial distribution with n = 5 and p = 1/4

(i) P (all give cards are spades) = P(X = 5)

(ii) P(only 3 cards are spades) = P(X = 3)

(iii) P (none is a spade) = P(x = 0)

**5. The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. What is the probability that out of 5 such bulbs(i) none(ii) not more than one(iii) more than one(iv) at least onewill fuse after 150 days of use.**

**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

∴ q = 1 - p = 1 - 0.05 = 0.95

X has a binomial distribution with n = 5 and p = 0.05

(i) P (none) = P(x = 0)

(ii) P(not more than one) = P(X ≤ 1)

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

(iii) P (more than 1) = P(X > 1)

= 1 - P(X ≤ 1)

= 1 - P(not more than 1)

= 1 - (0.95)^{4} × 1.2

(iv) P (at least one) = P(X ≥ 1)

**6. 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

∴ q = 1 - p = 1 - 1/10 = 9/10

**7. 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.**

**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.

∴ p = 1/2

∴ q = 1 - p = 1 - 1/2 = 1/2

X has a binomial distribution with n = 20 and p = 1/2

**8. 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 ( x_{i}), x_{i} = 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

∴ q = 1 - p = 1 - 1/2 = 1/2

It can be seen that P(X = x) will be maximum, if ^{6}C_{x} will be maximum.

The value of ^{6}C_{x} is maximum. Therefore, for x = 3, P(X = x) is maximum.

Thus, X = 3 is the most likely outcome.

**9. 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

∴ q = 1 - p = 1 - 1/3 = 2/3

Clearly, X has a binomial distribution with n = 5 and p = 1/3

P(guessing more than 4 correct answers) = P(X ≥ 4)

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

**10. 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 in 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

∴ q = 1 - p = 1 - 1/100 = 99/100

(a) P(winning at least once) = P(X ≥ 1)

= 1 - P(X < 1)

= 1 - P(X = 0)

(b) P (winning exactly once) = P(X = 1)

(c) P (at least twice) = P(X ≥ 2)

= 1 - P(X < 2)

= 1 - P(X ≤ 1)

= 1 - [P(X = 0) + P(X = 1)]

= [1 - P(X = 0)] - P(X = 1)

**11. 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, p = 1/6

∴ q = 1 - p = 1 - 1/6 = 5/6

Clearly, X has the probability distribution with n = 7 and p = 1/6

**12. 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

∴ q = 1 - p = 1 - 1/6 = 5/6

Clearly, X has a binomial distribution with n = 6

**13. 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 p = 10% = 10/100 = 1/10

∴ q = 1- p = 1 - 1/10 = 9/10

**14. 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)^{4} (C) (9/10)^{5} (D) 9/10**

**Solution**

The repeated selections of defective bulbs from a box are Bernoulli trials. Let X denote the number of defective bulbs out of a sample of 5 bulbs.

Probability of getting a defective bulb, p = 10/100 = 1/10

∴ q = 1 - p = 1 - 1/10 = 9/10

Clearly, X has a binomial distribution with n = 5 and p = 1/10

The correct answer is C.

**15. 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) ^{5}C_{4} (4/5)^{4} 1/5**

**(B) (4/5) ^{4} 1/5**

**(C) ^{5}C_{1} 1/5(4/5)^{4}**

**(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

∴ p = 1 - q = 1 - 1/5 = 4/5

Clearly, X has a binomial distribution with n = 5 and p = 4/5

Therefore, the correct answer is A.