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

NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.5 – Free PDF Download

Exercise 13.5 of NCERT Solutions for Class 12 Maths Chapter 13 – Probability is based on the following topics:

1. Bernoulli Trials and Binomial Distribution
   1. Bernoulli trials: Trials of a random experiment is called Bernoulli trials, if they satisfy the following conditions:
      1. There should be a finite number of trials.
      2. The trials should be independent.
      3. Each trial has exactly two outcomes: success or failure.
      4. The probability of success remains the same in each trial.
   2. Binomial distribution

The problems of this exercise are based on the topics listed above and hence, solving these questions will help gain an upper hand on these topics.

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

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Access Other Exercises of Class 12 Maths Chapter 13

Exercise 13.1 Solutions 17 Questions

Exercise 13.2 Solutions 18 Questions

Exercise 13.3 Solutions 14 Questions

Exercise 13.4 Solutions 17 Questions

Miscellaneous Exercise On Chapter 13 Solutions 10 Questions

Access Answers to NCERT Class 12 Maths Chapter 13 Exercise 13.5

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:

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)

Thus, q = 1 – p = ½

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

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

Solution:

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.

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

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 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

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 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

= 1/1024

5. 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.

Solution:

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

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 us assume that number of balls with digit marked as zero among the experiment of 4 balls drawn simultaneously is 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

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 us assume that the number of correctly answered questions out of twenty questions is 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

8. 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)

Solution:

Given X is any random variable whose binomial distribution is B (6, ½)

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.

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:

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

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 win a prize
(a) At least once
(b) Exactly once
(c) At least twice?

Solution:

(a) Let X represent 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

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

Solution:

Let us assume X represents 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

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

Solution:

Let us assume X represents 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

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:

Let us assume X represents 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

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)⁵
C. (9/10)⁵
D. 9/10 

Solution:

C. (9/10)⁵

Explanation:

Let us assume X represents 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

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. ⁵C₄ 1/5 (4/5)⁴
B. (4/5)⁴ (1/5)
C. ⁵C₁ 1/5 (4/5)⁴
D. None of these

Solution:

A. ⁵C₄ 1/5 (4/5)⁴

Explanation:

Let us assume X represents 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

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NCERT Solutions for Class 12 Maths

NCERT Solutions for Class 12

NCERT Solutions