Class 12 RD Sharma Solutions - Chapter 33 Binomial Distribution - Exercise 33.2 | Set 2

Question 15.  A dice is thrown thrice. A success is 1 or 6 in a throw. Find the mean and variance of the number of successes.

Solution:

Let p denote the success and q denote failure of an event.

Now, the sample space when a dice is thrown is given by S = {1, 2, 3, 4, 5, 6}

Hence, p = 2/6 = 1/3 and q = 1 - 1/3 = 2/3

Therefore, Mean = np = 3 × 1/3 = 1 and Variance = npq = 1 × 2/3 = 2/3

Question 16. If a random variable X follows a binomial distribution with mean 3 and variance 3/2, find P (X ≤ 5).

Solution:

We are given mean (np) = 3 and variance (npq) = 3/2.

Solving for the value of q, 

q = 1/2, hence we can conclude p = 1 - 1/2 = 1/2

Now putting the value of p in relation, np = 3, we get n = 6

We know that a binomial distribution follows the relation:

P(X = r) = ⁿC_r p^r(q)^n-r

Therefore, in this case P(X = r) = ⁶C_r (1/2)^r(1/2)^6-r

P(X = r) = ⁶C_r (1/2)⁶

We are required to calculate the value for P(X ≤ 5) = 1 - P(X = 6)

P(X ≤ 5) = 1 - ⁶C_r (1/2)⁶

P(X ≤ 5) = 1 - (1/64) 

P(X ≤ 5) = 63/64 

Question 17. If X follows a binomial distribution with mean 4 and variance 2, find P(X ≥ 5).

Solution:

We are given mean (np) = 4 and variance (npq) = 2.

Solving for the value of q, npq/np = 2/4 

q = 1/2, hence we can conclude p = 1 - 1/2 = 1/2

Now putting the value of p in relation, np = 4, we get n = 8

We know that a binomial distribution follows the relation: P(X = r) = ⁿC_r p^r(q)^n-r

Therefore, in this case P(X = r) = ⁸C_r (1/2)^r(1/2)^8-r

P(X = r) = ⁸C_r (1/2)⁸

We are required to calculate the value 

P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)

P(X ≥ 5) = ⁸C₅ (1/2)⁸ + ⁸C₆ (1/2)⁸ + ⁸C₇ (1/2)⁸ + ⁸C₈ (1/2)⁸

P(X ≥ 5) = (1/2)⁸[⁸C₅ + ⁸C₆ + ⁸C₇ + ⁸C₈]

P(X ≥ 5) = (56 + 28 + 8 + 1)/256

P(X ≥ 5) = 93/256 

Question 18. The mean and variance of a binomial distribution are 4/3 and 8/9 respectively. Find P(X ≥ 1).

Solution:

We are given mean (np) = 4 and variance (npq) = 2

Solving for the value of q, npq/np = 

q = 2/3, hence we can conclude p = 1 - 2/3 = 1/3

Now putting the value of p in relation, np = 4/3, we get n = 4

We know that a binomial distribution follows the relation: P(X = r) = ⁿC_r p^r(q)^n-r

Therefore, in this case P(X = r) = ⁴C_r (1/3)^r(2/3)^4-r

We are required to calculate the value for P(X ≥ 1) = 1 - P(X = 0)

P(X ≥ 1) = 1 - ⁴C₀ (1/3)⁰(2/3)⁴

P(X ≥ 1) = 1 - 16/81

P(X ≥ 1) = 65/81

Question 19. If the sum of the mean and variance of a binomial distribution for 6 trials is 10/3, find the distribution.

Solution:

Given n = 6 and np + npq = 10/3

np (1 + q) = 10/3

6p (1 + 1 - p) = 10/3

12p - 6p² = 10/3

18p² - 36p + 10 = 0

Solving for the value of p we will get p = 1/3 or p = 5/3. 

Since, the value of p cannot exceed 1, we will consider p = 1/3.

Therefore, q = 1 - 1/3 = 2/3

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

P(x = r) = ⁶C_r (1/3)^r(2/3)^6-r for r = 0,1,2,....,6

Question 20. A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of the number of successes and, hence, find its mean.

Solution:

We are given n = 4 and 

a doublet in the throw of a dice occurs when we get (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)

Therefore, the probability of success, p = 6/36 = 1/6, so q = 1 - 1/6 = 5/6

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

P(x = r) = ⁴C_r (1/6)^r(5/6)^4-r for r = 0, 1, 2, 3, 4

Hence, the probability distribution is given as:

X	0	1	2	3	4
P(X)	625/1296	500/1296	150/1296	20/1296	1/1296

Mean = 0 × (625/1296) + 1 × (500/1296) + 2 × (150/1296) + 3 × (20/1296) + 0 × (1/1296)

= 864/ 1296

= 2/3

Question 21. Find the probability distribution of the number of doublets in three throws of a pair of dice and find its mean.

Solution:

We are given n = 3 and 

a doublet in the throw of a dice occurs when we get (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)

Therefore, the probability of success, p = 6/36 = 1/6, so q = 1 - 1/6 = 5/6

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

P(x = r) = ³C_r (1/6)^r(5/6)^3-r for r = 0, 1, 2, 3

Hence, the probability distribution is given as:

X	0	1	2	3
P(X)	125/216	75/216	15/216	1/216

Mean = 0 × (125/216) + 1 × (75/216) + 2 × (15/216) + 3 × (1/216) 

= 108/216

= 1/2

Question 22. From a lot of 15 bulbs which include 5 defective, a sample of 4 bulbs is drawn one by one with replacement. Find the probability distribution of number of defective bulbs. Hence, find the mean of the distribution.

Solution:

Total number of bulbs = 15 and total defective bulbs = 5 

Thus, the probability of getting one defective bulb with replacement, p = 5/15 = 1/3

 Hence, q = 1 - 1/3 = 2/3.

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

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

Hence, the probability distribution is given as:

X	0	1	2	3	4
P(X)	16/81	32/81	24/81	8/81	1/81

Mean = 0 × (16/81) + 1 × (32/81) + 2 × (24/81) + 3 × (8/81) + 4 × (1/81) 

= 108/81

= 4/3

Question 23. A die is thrown three times. Let X be 'the number of twos seen'. Find the expectation of X.

Solution:

We are given the number of throws, n = 3

Let p denote the probability of getting a 2 in the throw of a dice, then p = 1/6

Therefore, we can conclude 1 = 1 - 1/6 = 5/6

Now, the expectation of X denotes mean therefore, E(X) = np = 3 × 1/6 = 1/2

Question 24. A die is tossed twice. A 'success' is getting an even number on a toss. Find the variance of the number of successes.

Solution:

We are given the number of times the coin is tossed, n = 2

Let p denote the probability of getting even number on dice upon throwing which is a success.

Thus, p = 3/6 = 1/2, therefore we can conclude q = 1 - p = 1 - 1/2 = 1/2

Now, the variance is given by npq.

Variance = 2 × 1/2 × 1/2 = 1/2

Question 25. Three cards are drawn successively with replacement from a well-shuffled pack of 52 cards. Find the probability distribution of the number of spades. Hence, find the mean of the distribution. 

Solution:

Number of cards drawn with replacement, n = 3

p = Probability of getting a spade card upon withdrawal = 13/52 = 1/4

Thus, we can conclude, q = 1 - 1/4 = 3/4

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

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

Hence, the probability distribution is given as:

X	0	1	2	3
P(X)	27/64	27/64	9/64	1/64

Mean = 0 × (27/64) + 1 × (27/64) + 2 × (9/64) + 3 × (1/64) 

= (27 + 18 + 3)/64

= 48/64 

= 3/4

Question 26. An urn contains 3 white and 6 red balls. Four balls are drawn one by one with replacement from the urn. Find the probability distribution of the number of red balls drawn. Also, find mean and variance of the distribution.

Solution:

Let p denote the probability of drawing a red ball which is considered a success, p = 6/9 = 2/3

And the probability of drawing a white ball which is considered a failure, q = 3/9 = 1/3

We have to draw four balls, so n = 4.

Hence, the mean of the probability distribution = np = 4 × 2/3 = 8/3

And variance = npq = 8/3 × 1/3 = 8/9

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

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

Hence, the probability distribution is given as:

X	0	1	2	3	4
P(X)	1/81	8/81	24/81	32/81	16/81

Summary

Exercise 33.2 (Set 2) 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 box of 20 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 7 heads?

3. The probability of a student passing a test is 0.8. If 5 students take the test, what is the probability that exactly 4 students pass?

4. In a group of 12 people, the probability of a person having a certain disease is 0.25. Find the probability that exactly 3 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 2 of them are 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 12 light bulbs are produced, find the probability that at least 10 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.