Class 12 RD Sharma Solutions Chapter 33 Binomial Distribution Exercise 33.1 | Set 1 (original) (raw)

Last Updated : 23 Jul, 2025

The Binomial Distribution is a fundamental concept in probability theory and statistics particularly useful in scenarios involving discrete outcomes. It provides a way to model the number of successes in a fixed number of independent trials where each trial has the same probability of success. In this article, we will delve into Exercise 33.1 | Set 1 from Chapter 33 of the RD Sharma Class 12 textbook solving the various problems related to the Binomial Distribution to enhance understanding and application.

Binomial Distribution

The Binomial Distribution describes the probability of having exactly k successes in the n independent Bernoulli trials each with the same probability p of success. The probability mass function of the Binomial Distribution is given by:

P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}

where:

Question 1. There are 6% defective items in a large bulk of items. Find the probability that a sample of 8 items will include not more than one defective item.

**Solution:

Let us consider X be the number of defective items in a sample of 8 items.

So, a binomial distribution follows by x with n = 8.

and Probability of getting a defective item(p) = 0.06

The probability of getting a defective item (1 - p) = 0.94

So, P(X = r) = 8Cr (0.06)r (0 . 94 )8 - r, where r = 0, 1, 2, 3, . . . 8

So, the required probability is

= P (X < 1)

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

= 8C0 (0.06)0 (0.94)8-0 + 8C1 (0.06)1 (0.94 )8-1

= (0.94)8 + 8 (0.06) (0.94)7

= (0.94)7 {0.94 + 0.48}

= 1.42 (0.94)7

Question 2. A coin is tossed 5 times. What is the probability of getting at least 3 heads?

**Solution:

Let us consider X be the number of heads in 5 tosses.

So, a binomial distribution follows by x with n = 5

Probability of getting a head(p) = 1/2

And q = 1 - p = 1 - 1/2 = 1/2

P(X = r) = 5Cr (1/2)r (1/2)5-r , where r = 0, 1, 2 . . . 5

So, the required probability is

= P(X > 3)

= P(X = 3) + P(X = 4) + P(X = 5)

= 5C3 (1/2)3 (1/2)5-3 + 5C4 (1/2)1 (1/2)5-1 + 5C5 (1/2)5 (1/2)5-0

= 10 (1/2)5 + 5 (1/2)5 + 1 (1/2)5

= (1/2)5 (10 + 5 + 1)

= (1/2)5 (16)

= 1/2

Question 3. A coin is tossed 5 times. What is the probability that tail appears an odd number of times?

**Solution:

Let us consider X be the number of tails when a coin is tossed 5 times.

So, a binomial distribution follows by x with n = 5

Probability of getting head(p) = 1/2.

Also, q = 1 - p = 1/2

P(X = r) = 5C3 (1/2)r (1/2)n-r = 5Cr (1/2)5

So, the required probability is

= P(X = 1) + P(X = 3) + P(X = 5)

= 5C1 (1/2)5 + 5C3 (1/2)5 + 5C5 (1/2)5

= (1/2)5 [5 + 10 + 1]

= 16/32

= 1/2

Question 4. A pair of dice is thrown 6 times. If getting a total of 9 is considered a success, what is the probability of at least 5 successes?

**Solution:

Let us consider X be the number of successes in 6 throws of the two dice.

So the probability of success is equal to the probability of getting a total of 9

= Probability of getting (3, 6), (4, 5), (5, 4), (6, 3) from 36 outcomes

Here, p = 4/36 = 1/9

Also q = 1 - p = 8/9 and n = 6

Now, a binomial distribution X follows with n = 6, p = 1/9, and q = 8/9

P(X = r) = 6Cr (1/9)r (8/9)6-r

The required probability is

= P(X > 5)

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

= ^{6}{}{C}_5 \left( \frac{1}{9} \right)^5 \left( \frac{8}{9} \right)^{6 - 5} + ^{6}{}{C}_6 \left( \frac{1}{9} \right)^6 \left( \frac{8}{9} \right)^{6 - 6}

= \frac{6(8) + 1}{9^6}

= 49/96

Question 5. A fair coin is tossed 8 times, find the probability:

(i) of exactly 5 heads.

**Solution:

Let us consider X be the number of heads obtained when a fair is tossed 8 times.

Now, a binomial distribution X follows with n = 8.

Here, p = 1/2 and q = 1 - 1/2 = 1/2

So, P(X = r) = ^8 C_r \left( \frac{1}{2} \right)^{8 - r} \left( \frac{1}{2} \right)^r

= 8Cr (1/2)8 , where r = 0, 1, 2, . . . , 8

So, the probability of obtaining exactly 5 heads is

P(X = 5) = 8C5 (1/2)8

= 56/256

= 7/32

(ii) of at least six heads.

**Solution:

Let us consider X be the number of heads obtained when a fair is tossed 8 times.

Now, a binomial distribution X follows with n = 8.

Here, p = 1/2 and q = 1 - 1/2 = 1/2

So, P(X = r) = ^8 C_r \left( \frac{1}{2} \right)^{8 - r} \left( \frac{1}{2} \right)^r

= 8Cr (1/2)8 , where r = 0, 1, 2, . . . , 8

So, the probability of obtaining exactly 5 heads is

P(X = 5) = 8C5 (1/2)8

= 56/256

= 7/32

(iii) of at most six heads.

**Solution:

Let us consider X denotes the number of heads obtained when a fair is tossed 8 times.

Now, a binomial distribution X follows with n = 8.

Here p = 1/2 and q = 1 - 1/2 = 1/2

P(X = r) = ^8 C_r \left( \frac{1}{2} \right)^{8 - r} \left( \frac{1}{2} \right)^r

= 8Cr (1/2)8 , r = 0, 1, 2, . . . , 8

so the probability of obtaining at most 6 heads is

P (X < 6) = 1 - [P(X = 7) + P(X = 8)]

= 1 - \left[ {}^8 C_7 \left( \frac{1}{2} \right)^8 +^8 C_8 \left( \frac{1}{2} \right)^8 \right]

= 1 - (8/256 + 1/256)

= 1 - 9/256

= 247/256

Question 6. Find the probability of 4 turning up at least once in two tosses of a fair die.

**Solution:

Let us consider X denotes the probability of getting 4 in two tosses of a fair die.

Now, a binomial distribution X follows with n = 2.

Here, p = 1/6 and q = 5/6

P(X = r) = ^{2}{}{C}_r \left( \frac{1}{6} \right)^r \left( \frac{5}{6} \right)^{2 - r}

So, the probability of obtaining 4 at least once is

P(X > 1) = 1 - P(X = 0)

= 1 - ^{2}{}{C}_0 \left( \frac{1}{6} \right)^0 \left( \frac{5}{6} \right)^{2 - 0}

= 1 - 25/36

= 11/36

Question 7. A coin is tossed 5 times. What is the probability that head appears an even number of times?

**Solution:

Let us consider X denotes the number of heads that appear when a coin is tossed 5 times.

Now, a binomial distribution X follows with n = 5

Here p = q = 1/2.

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{5 - r}

= 5Cr (1/2)5

P (head appears an even number of times) = P(X = 0) + P(X = 2) + P(X = 4)

= ^{5}{}{C}_0 \left( \frac{1}{2} \right)^5 + ^{5}{}{C}_2 \left( \frac{1}{2} \right)^5 + ^{5}{}{C}_4 \left( \frac{1}{2} \right)^5

= \frac{1 + 10 + 5}{2^5}

= 16/32

= 1/2

Question 8. The probability of a man hitting a target is 1/4. If he fires 7 times, what is the probability of his hitting the target at least twice?

**Solution:

Let us consider X denotes the number of times the target is hit.

Now, a binomial distribution X follows with n = 7.

Here, p = 1/4 and q = 3/4.

P(X = r) = ^{7}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{7 - r}

P(hit the target at least twice) = P(X > 2)

= 1 - {P(X = 0) + P(X = 1)}

= 1 -^{7}{}{C}_0 \left( \frac{1}{4} \right)^0 \left( \frac{3}{4} \right)^{7 - 0} - ^{7}{}{C}_1 \left( \frac{1}{4} \right)^1 \left( \frac{3}{4} \right)^{7 - 1}

= 1 - (3/4)7 - 7 (1/4) (3/4)6

= 1 - 1/16384(2187 + 5103)

= 1 - 3645/8192

= 4547/8192

Question 9. Assume that on average one telephone number out of 15 called between 2 P.M. and 3 P.M. on weekdays is busy. What is the probability that if six randomly selected telephone numbers are called, at least 3 of them will be busy?

**Solution:

Let us consider X denotes the number of busy calls for 6 randomly selected telephone numbers.

Now, a binomial distribution X follows with n = 6.

Here, p = one out of 15 = 1/15

And q = 1 - 1/15 = 14/15

P(X = r) = ^{6}{}{C}_r \left( \frac{1}{15} \right)^r \left( \frac{14}{15} \right)^{6 - r}

So, the probability that at least 3 of them are busy is

P(X > 3) = 1 - {P(X = 0) + P(X = 1) + P(X = 2)}

= 1 - \left\{ ^{6}{}{C}_0 \left( \frac{1}{15} \right)^0 \left( \frac{14}{15} \right)^{6 - 0} + ^{6}{}{C}_1 \left( \frac{1}{15} \right)^1 \left( \frac{14}{15} \right)^{6 - 1} + ^{6}{}{C}_2 \left( \frac{1}{15} \right)^2 \left( \frac{14}{15} \right)^{6 - 2} \right\}

= 1 - \left\{ \left( \frac{14}{15} \right)^6 + \frac{6}{15} \left( \frac{14}{15} \right)^5 + \frac{1}{15} \left( \frac{14}{15} \right)^4 \right\}

Question 10. If getting 5 or 6 in a throw of an unbiased die is a success and the random variable X denotes the number of successes in six throws of the die, find P (X ≥ 4).

**Solution:

Let us consider X be the number of successes. That is getting 5 or 6 in a throw of die in 6 throws.

Now, a binomial distribution X follows with n = 6

Here, p = of getting 5 or 6 = 1/6 + 1/6 = 1/3

And q = 1 - p = 2/3

P(X = r) = ^{6}{}{C}_r \left( \frac{1}{3} \right)^r \left( \frac{2}{3} \right)^{6 - r}

P(X > 4) = P(X = 4) + P(X = 5) + P(X = 6)

= ^{6}{}{C}_4 \left( \frac{1}{3} \right)^4 \left( \frac{2}{3} \right)^{6 - 4} +^{6}{}{C}_5 \left( \frac{1}{3} \right)^5 \left( \frac{2}{3} \right)^{6 - 5} + ^{6}{}{C}_6 \left( \frac{1}{3} \right)^6 \left( \frac{2}{3} \right)^{6 - 6}

= (1/36) (60 + 12 + 1)

= 73/729

Question 11. Eight coins are thrown simultaneously. Find the chance of obtaining at least six heads.

**Solution:

Let us consider X denotes the number of heads in tossing 8 coins.

Now, a binomial distribution X follows with n = 8.

Here, p = 1/2 and q = 1/2

P(X = r) = ^{8}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{8 - r}

= 8Cr (1/2)8

So, the probability of getting at least 6 heads is

P(X > 6) = P(X = 6) + P(X = 7) + P(X = 8)

= ^{8}{}{C}_6 \left( \frac{1}{2} \right)^8 + ^{8}{}{C}_7 \left( \frac{1}{2} \right)^8 + ^{8}{}{C}_8 \left( \frac{1}{2} \right)^8

= 1/28(28 + 8 + 1)

= 37/256

Question 12. Five cards are drawn successively with replacement from a well-shuffled pack of 52 cards.

(i) What is the probability that all the five cards are spades?

**Solution:

Let us consider X be the number of spade cards when 5 cards are drawn with replacement.

Now, a binomial distribution X follows with n = 5.

p = 13/52 = 1/4 and q = 1 - p = 3/4

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{5 - r}

P(All cards are spades) = P(X = 5)

= ^{5}{}{C}_5 \left( \frac{1}{4} \right)^5 \left( \frac{3}{4} \right)^0

= 1/1024

(ii) What is the probability that only 3 cards are spades?

**Solution:

Let us consider X be the number of spade cards when 5 cards are drawn with replacement.

Now, a binomial distribution X follows with n = 5.

Here, p = 13/52 = 1/4 and q = 1 - p = 3/4

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{5 - r}

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

= ^{5}{}{C}_3 \left( \frac{1}{4} \right)^3 \left( \frac{3}{4} \right)^2

= (1/1024) (90)

= 45/512

(iii) What is the probability that none is a spade?

**Solution:

Let us consider X be the number of spade cards when 5 cards are drawn with replacement.

Now, a binomial distribution X follows with n = 5.

Here p = 13/52 = 1/4 and q = 1 - p = 3/4.

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{5 - r}

P(none is a spade) = P(X = 0)

= ^{5}{}{C}_0 \left( \frac{1}{4} \right)^0 \left( \frac{3}{4} \right)^5

= 243/1024

Question 13. A bag contains 7 red, 5 white, and 8 black balls. Four balls are drawn one by one with replacement.

(i) What is the probability that none is white?

**Solution:

Let us consider X denotes the number of white balls drawn when 4 balls are drawn with replacement.

Now, a binomial distribution X follows with n = 4.

Here, the probability for a white ball(p) = 5/20 = 1/4

And q = 1 - p = 3/4

P(X = r) = ^{4}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{4 - r}

So, the probability that none is white is

P(X = 0) = ^{4}{}{C}_0 \left( \frac{1}{4} \right)^0 \left( \frac{3}{4} \right)^{4 - 0}

= 81/256

(ii) What is the probability that none is white?

**Solution:

Let us consider X denotes the number of white balls drawn when 4 balls are drawn with replacement.

Now, a binomial distribution X follows with n = 4.

Here, the probability for a white ball(p) = 5/20 = 1/4

And q = 1 - p = 3/4

P(X = r) = ^{4}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{4 - r}

so, the probability that none is white is

P(X = 0) = ^{4}{}{C}_0 \left( \frac{1}{4} \right)^0 \left( \frac{3}{4} \right)^{4 - 0}

= 81/256

(iii) What is the probability that any two are white?

**Solution:

Let us consider X denotes the number of white balls drawn when 4 balls are drawn with replacement.

Now, a binomial distribution X follows with n = 4.

So, the probability for a white ball(p) = 5/20 = 1/4

And q = 1 - p = 3/4

P(X = r) = ^{4}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{4 - r}

So, the probability that any two are white is

P(X = 2) = ^{4}{}{C}_2 \left( \frac{1}{4} \right)^2 \left( \frac{3}{4} \right)^{4 - 2}

= 54/256

= 27/128

Question 14. A box contains 100 tickets, each bearing one of the numbers from 1 to 100. If 5 tickets are drawn successively with replacement from the box, find the probability that all the tickets bear numbers divisible by 10.

**Solution:

Let us consider X denotes the variable representing number on the ticket bearing a number divisible by 10 out of the 5 tickets drawn.

Now, a binomial distribution X follows with n = 5.

Here, the probability of getting a ticket bearing number divisible by 10(p) = 10/100

= 1/10

And q = 1 - p = 9/10

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{10} \right)^r \left( \frac{9}{10} \right)^{5 - r}

So, the probability that all the tickets bear numbers divisible by 10 is

P(X = 5) = ^{5}{}{C}_5 \left( \frac{1}{10} \right)^5 \left( \frac{9}{10} \right)^{5 - 5}

= (1/10)5 (9/10)0

= (1/10)5

Question 15. A bag contains 10 balls, each marked with one of the digits from 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 consider X denotes the number of balls marked with the digit 0 when 4 balls are drawn successfully with replacement.

Now, a binomial distribution X follows with n = 4.

So, the probability that a ball randomly drawn bears digit 0(p) = 1/10

And q = 1 - p = 9/10

P(X = r) = ^{4}{}{C}_r \left( \frac{1}{10} \right)^r \left( \frac{9}{10} \right)^{4 - r}

P(none bears the digit 0) = P(X = 0)

= ^{4}{}{C}_0 \left( \frac{1}{10} \right)^0 \left( \frac{9}{10} \right)^{4 - 0}

= (9/10)4

Question 16. In a large bulk of items, 5 percent of the items are defective. What is the probability that a sample of 10 items will include not more than one defective item?

**Solution:

Let us consider X be the number of defective items in a sample of 10 items.

Now, a binomial distribution X follows with n = 10.

And the probability of defective items(p) = 5% = 0.05

And q = 1 - p = 0.95

P(X = r) = 10Cr (0. 05)r (0.95)10-r

So, the probability(sample of 10 items will include not more than one defective item) is

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

= 10C0 (0.05)0 (0.95 ){0-0 + 10C1 (0.05)1 (0.95)10-1

= (0.95)9 (0.95 + 0.5)

= 1.45 (0.95)9

Question 17. The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05.

(i) Find the probability that out of 5 such bulbs none will fuse after 150 days of use.

**Solution:

Let us consider X be the number of bulbs that fuse after 150 days.

Now, a binomial distribution X follows with n = 5.

Here, p = 0.05 and q = 0.95

Or p = 1/20 and q = 19/20

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{20} \right)^r \left( \frac{19}{20} \right)^{5 - r}

So, the probability (none will fuse after 150 days of use) is

P(X = 0) = ^{5}{}{C}_0 \left( \frac{1}{20} \right)^0 \left( \frac{19}{20} \right)^{5 - 0}

= (19/20)5

(ii) Find the probability that out of 5 such bulbs, not more than one will fuse after 150 days of use.

**Solution:

Let us consider X denotes the number of bulbs that fuse after 150 days.

Now, a binomial distribution X follows with n = 5.

Here p = 0.05 and q = 0.95.

Or p = 1/20 and q = 19/20

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{20} \right)^r \left( \frac{19}{20} \right)^{5 - r}

So, the probability (not more than 1 will fuse after 150 days of use) is

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

= \left( \frac{19}{20} \right)^5 + 5 C_1 \left( \frac{1}{20} \right)^1 \left( \frac{19}{20} \right)^{5 - 1}

= \left( \frac{19}{20} \right)^4 \left\{ \frac{19}{20} + \frac{5}{20} \right\}

= (6/5) (19/20)4

(iii) Find the probability that out of 5 such bulbs more than one will fuse after 150 days of use

**Solution:

Let us consider X denotes the number of bulbs that fuse after 150 days.

Now, a binomial distribution X follows with n = 5.

Here p = 0.05 and q = 0.95

Or p = 1/20 and q = 19/20

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{20} \right)^r \left( \frac{19}{20} \right)^{5 - r}

So, the probability(more than one will fuse after 150 days of use) is

P(X > 1) = 1 - P(X < 1)

= 1 - (6/5) (19/20)4

(iv) Find the probability that out of 5 such bulbs at least one will fuse after 150 days of use.

**Solution:

Let us consider X denotes the number of bulbs that fuse after 150 days.

Now, a binomial distribution X follows with n = 5.

Here p = 0.05 and q = 0.95

Or p = 1/20 and q = 19/20

P(X = r) = ^{5}{}{C}_r \left( \frac{1}{20} \right)^r \left( \frac{19}{20} \right)^{5 - r}

So, the probability(at least one will fuse after 150 days of use) is

P(X > 1) = 1 - P(X = 0)

= 1 - (19/20)5

Question 18. 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?

**Solution:

Let us consider X denotes the number of people that are right-handed in the sample of 10 people.

Now, a binomial distribution X follows with n = 10.

Here p = 90 % = 90/100 = 0.9

And q = 1 - p = 0.1

P(X = r) = 10Cr (0.9)r (0.1)10-r

So, the probability that at most 6 are right - handed is

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)

= 1 - {P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)}

= 1 - \sum^{10}_{r = 7}{^{10}{}{C}_r} (0 . 9 )^r (0 . 1 )^{10 - r}

Summary

Exercise 33.1 (Set 1) in Chapter 33 (Binomial Distribution) of RD Sharma's Class 12 mathematics textbook focuses on introducing the concepts of binomial distribution and solving problems related to finding the probability of a specific number of successes in a given number of trials. The exercise covers topics such as understanding the characteristics of the binomial distribution, calculating the probability of a specific outcome, and applying the binomial probability formula to solve real-world problems. This set of problems helps students develop a solid foundation in the binomial distribution and its applications in probability and statistics.

Practice Questions

1. In a box of 12 light bulbs, the probability of a bulb being defective is 0.15. Find the probability that exactly 3 bulbs are defective.

2. A fair coin is tossed 10 times. What is the probability of getting exactly 6 heads?

3. The probability of a student passing a test is 0.8. If 6 students take the test, what is the probability that exactly 5 students pass?

4. In a group of 14 people, the probability of a person having a certain disease is 0.25. Find the probability that exactly 4 people have the disease.

5. A manufacturer produces electronic components, and the probability of a component being defective is 0.1. If 8 components are randomly selected, what is the probability that exactly 1 component is defective?

6. In a quality control process, the probability of a product being defective is 0.18. If 15 products are inspected, what is the probability that at most 3 products are defective?

7. A company that manufactures light bulbs has a 95% success rate. If 10 light bulbs are produced, find the probability that at least 9 of them are successful.

8. In a group of 8 students, the probability of a student passing an exam is 0.6. Find the probability that exactly 5 students pass the exam.

9. The probability of a person buying a particular product is 0.4. If 6 people are randomly selected, what is the probability that at least 3 of them buy the product?

10. A biased coin is tossed 15 times, and the probability of getting a head on each toss is 0.7. Find the probability of getting exactly 12 heads.