Class 12 RD Sharma Solutions- Chapter 33 Binomial Distribution - Exercise 33.2 | Set 1

Exercise 33.2 (Set 1) in RD Sharma's Class 12 Mathematics book focuses on the application of the binomial distribution, a fundamental concept in probability theory. In this exercise, students will explore various problems that involve calculating probabilities in scenarios where there are only two possible outcomes—success or failure. The exercise is designed to enhance students' understanding of how to use the binomial distribution formula effectively and to solve real-world problems related to probability. This chapter is crucial for building a strong foundation in probability, which is an important topic for higher studies in mathematics and statistics.

Question 1. Can the mean of a binomial distribution be less than its variance?

Solution:

Let np be the mean and npq be the variance of a binomial distribution.

So, 

Mean - Variance = np - npq

Mean - Variance = np (1 - q)

Mean - Variance = np.p

Mean - Variance = np²

Since n can never be a negative number and 

p² will always be a positive number, thus np² > 0 

Then, 

Mean - Variance > 0

Mean > Variance

Hence, a mean of a binomial distribution can never be less than its variance.

Question 2. Determine the binomial distribution whose mean is 9 and variance 9/4.

Solution:

We are given mean(np) = 9 and variance(npq) = 9/4.

Solving for the value of q, we will get 

q = \frac{\frac{9}{4}}{9} = \frac{1}{4}

We know, the relation p + q = 1

So, p = 1 - (1/4) = 3/4          -(1) 

Since, np = 9

So, put the value of p from equation(1), we get

n.(3/4) = 9

n = 12

Now, a binomial distribution is given by the relation: ⁿC_r p^r(q)^n-r

P(x = r) = ¹²C_r(3/4)^r(1/4)^12-r for r = 0,1,2,3,4,....,12

Question 3. If the mean and variance of a binomial distribution are respectively 9 and 6, find the distribution.

Solution:

We are given mean, np = 9 and variance npq = 6.

Solving for the value of q, we will get 

q = 6/9 = 2/3 

We know, the relation p + q = 1

p = 1 - (2/3) = 1/3          -(1)  

Since, np = 9

So, put the value of p from equation(1), we get

n.(1/3) = 9  

n = 27

Now, a binomial distribution is given by the relation: ⁿC_r p^r(q)^n-r

P(x = r) = ²⁷C_r(1/3)^r(2/3)^27-r for r = 0,1,2,3,4,...,27

Question 4. Find the binomial distribution when the sum of its mean and variance for 5 trials is 4.8.

Solution:

Given n = 5 and np + npq = 4.8

np (1 + q) = 4.8

5p (1 + 1 - p) = 4.8

10p -5p² = 4.8

50p² - 100p + 48 = 0

Solving for the value of p we will get 

p = 6/5 or p = 4/5

Since, the value of p cannot exceed 1, we will consider p = 4/5.

Therefore, q = 1 - 4/5 = 1/5

Now, a binomial distribution is given by the relation: ⁿC_r p^r(q)^n-r

P(x = r) = ⁵C_r(4/5)^r(1/5)^5-r for r = 0,1,2,....,5

Question 5. Determine the binomial distribution whose mean is 20 and variance 16.

Solution:

We are given mean, np = 20 and variance npq = 16.

Solving for the value of q, we will get 

q = 16/20 = 4/5 

We know, the relation p + q = 1

p = 1 - 4/5 = 1/5          -(1)  

Since, np = 20

So, put the value of p from equation(1), we get

n.(1/5) = 20

n = 100

Now, a binomial distribution is given by the relation: ⁿC_r p^r(q)^n-r

P(x = r) = ¹⁰⁰C_r(1/5)^r(4/5)^100-r for r = 0,1,2,3,4,...,100

Question 6. In a binomial distribution, the sum and product of the mean and the variance are 25/3 and 50/3  respectively. Find the distribution.

Solution:

We are given sum, np + npq = 25/3 

np(1 + q) = 25/3          -(1)  

Product, np x npq = 50/3          -(2)  

Dividing equation(2) by equation(1), we get 

 \frac{np(npq)}{np(1+q)} = (50/3) × 3/25

npq = 2 (1 + q)

np(1 - p) = 2(2 - p)

np = \frac{2(2-p)}{1-p}               -(3)  

On substituting the value of equation(3) in the relation np + npq = 25/3, we get

\frac{2(2-p)}{1-p} + \frac{2(2-p)}{1-p}.q  = 25/3

\frac{2(2-p)}{1-p} + \frac{2(2-p)}{1-p} . (1 - p) = 25/3

\frac{2(2-p)}{1-p} + \frac{2(2-p)}{1-p} (1 + 1 - p) = 25/3

\frac{2(2-p)}{1-p} (2 - p) = 25/3

6p²+ p - 1 = 0

On solving for the value of p, we will get p = 1/3, therefore q = 2/3 

Now, putting value of p and q in the relation np + npq = 25/3

n.(1/3)(1 + (2/3)) = 25/3

n = 15

Now, a binomial distribution is given by the relation: ⁿC_r p^r(q)^n-r

P(x = r) = ¹⁵C_r(1/3)^r(2/3)^15-r for r = 0,1,2,3,4,...,15

Question 7. The mean of a binomial distribution is 20 and the standard deviation 4. Calculate the parameters of the binomial distribution.

Solution:

We are given mean, np = 20          -(1) 

Standard deviation, √npq = 4

npq = 16           -(2) 

On dividing the equation (ii) by equation (i), we get

q = 4/5

Therefore, p = 1 - q = 1 - 4/5 = 1/5

Now, since np = 20

n = 20 x 5 = 100

Now, a binomial distribution is given by the relation: ⁿC_r p^r(q)^n-r

P(x = r) = ¹⁰⁰C_r(1/5)^r(4/5)^100-r for r = 0,1,2,3,4,...,100

Question 8. If the probability of a defective bolt is 0.1, find the (i) mean and (ii) standard deviation for the distribution of bolts in a total of 400 bolts.

Solution:

We are given n = 400 and q = 0.1, therefore p = 0.9

(i) Mean = np = 400 × 0.9 = 360

(ii) Standard Deviation = √npq =√(400 × 0.9 ×0.1) = 6

Question 9. Find the binomial distribution whose mean is 5 and variance 10/3.

Solution:

We are given mean, np = 5 and variance npq = 10/3.

Solving for the value of q, we will get 

q = \frac{\frac{10}{3}}{5}  = 2/3 

We know, the relation p + q = 1

p = 1 - (2/3) = 1/3 

Since, np = 5

n.(1/3) = 5

n = 15

Now, a binomial distribution is given by the relation: ⁿC_r p^r(q)^n-r

P(x = r) = ¹⁵C_r(1/3)^r(2/3)^15-r for r = 0,1,2,3,4,...,15

Question 10. If on an average 9 ships out of 10 arrive safely at ports, find the mean and S.D. of the ships returning safely out of a total of 500 ships.

Solution:

We are given n = 500,

p = 9/10 and thus q = 1/10

Therefore, mean = np = 500 × 0.9 = 450

Standard deviation = √npq = √(450 × 0.1) = 6.71 

Question 11. The mean and variance of a binomial variate with parameters n and p are 16 and 8, respectively. Find P(X = 0), P(X = 1), and P(X ≥ 2).

Solution:

We are given, mean (np) = 16 and variance (npq) = 8

q = 8/16 = 1/2

Therefore, p = 1 - 1/2 = 1/2

Putting the value of p in the relation, np = 16

n = 16 x 2 = 32

Now, a binomial distribution is given by the relation: ⁿC_r p^r(q)^n-r

P (x = r) = ³²C_r(1/2)^r(1/2)^32-r for r = 0,1,2,3,4,...,32

Now, P(X = 0) = ³²C₀(1/2)³² = (1/2)³²

Similarly, P(X = 1) = ³²C₁(1/2)¹(1/2)³¹ = 32 × (1/2)³¹

Also, P(X ≥ 2) = 1 - P(X = 0) - P(X = 1)

1 - (1/2)³² - 32 × (1/2)³²

1 - \frac{33}{2^{32}}

Question 12. In eight throws of a die, 5 or 6 is considered a success. Find the mean number of successes and the standard deviation.

Solution:

We are given n = 8 and p = 2/6 = 1/3 therefore q = 2/3

Now, mean = np = 8 ×(1/3) = 8/3 and

Standard deviation √npq = \sqrt{8×\frac{1}{3}×\frac{2}{3}} = 4/3 

Question 13. Find the expected number of boys in a family with 8 children, assuming the sex distribution to be equally probable.

Solution:

We are given n = 8 and the probability of having a boy or girl is equal, so p = 1/2 and q = 1/2

Therefore, the expected number of boys in a family = np = 8 × 0.5  = 4

Question 14. The probability that an item produced by a factory is defective is 0.02. A shipment of 10,000 items is sent to its warehouse. Find the expected number of defective items and the standard deviation.

Solution:

We are given n = 10,000, also p = 0.02 therefore, q = 1 - 0.02 = 0.98

Now, the expected number of defective items = np = 10000 × 0.02 = 200

The standard deviation = √npq = \sqrt{10000×0.02×0.98}  = √196 = 14

Summary

Exercise 33.2 (Set 1) in Chapter 33 (Binomial Distribution) of RD Sharma's Class 12 mathematics textbook focuses on applying the concepts of binomial distribution to solve various probability-related problems. The exercise covers topics such as finding the probability of a specific number of successes in a given number of trials, calculating the expected value and variance of a binomial random variable, and using the binomial distribution to solve real-world problems. This set of problems helps students develop a deeper understanding of the binomial distribution and its applications in probability and statistics.

Practice Questions

1. In a group of 15 students, the probability of a student passing a test is 0.7. Find the probability that exactly 10 students pass the test.

2. A fair coin is tossed 8 times. What is the probability of getting exactly 5 heads?

3. The probability of a person winning a game is 0.4. If 6 people play the game, find the probability that exactly 3 people win.

4. In a box of 25 light bulbs, the probability of a bulb being defective is 0.2. Find the probability that exactly 5 bulbs are defective.

5. A manufacturer produces electronic components, and the probability of a component being defective is 0.15. If 10 components are randomly selected, what is the probability that at most 2 of them are defective?

6. In a quality control process, the probability of a product being defective is 0.12. If 18 products are inspected, what is the probability that at least 3 products are defective?

7. A company that manufactures light bulbs has a 90% success rate. If 9 light bulbs are produced, find the probability that at least 8 of them are successful.

8. In a group of 10 students, the probability of a student passing an exam is 0.6. Find the probability that exactly 6 students pass the exam.

9. The probability of a person buying a particular product is 0.3. If 7 people are randomly selected, what is the probability that at least 4 of them buy the product?

10. A biased coin is tossed 12 times, and the probability of getting a head on each toss is 0.8. Find the probability of getting exactly 10 heads.