Page 280 - DCOM203_DMGT204_QUANTITATIVE_TECHNIQUES_I

Unit 13: Binomial Probability Distribution

Solution: Notes
Let the event that a man will be alive 30 years hence be termed as a success. Therefore, n = 5 and
2
p
3
0 1 5 242
2
1. P(r 1) = 1 - P(r = 0) 1 5 C 0
3 3 243
2. P(r 3) = P(r = 3) + P(r = 4) +P(r = 5)

3 2 4 1 5
5 2 1 5 2 1 5 2 64
C C C
3 4 5
3 3 3 3 3 81
Example: Ten percent of items produced on a machine are usually found to be defective.
What is the probability that in a random sample of 12 items (1) none, (2) one, (3) two, (4) at the
most two, (5) at least two items are found to be defective?
Solution:
Let the event that an item is found to be defective be termed as a success. Thus, we are given n =
12 and p = 0.1.
1. P r 0 12 C 0.1 0 0.9 12 0.2824
0

2. P r 1 12 C 0.1 1 0.9 11 0.3766
1
3. P r 2 12 C 0.1 2 0.9 10 0.2301
2
4. P r 2 = P(r = 0) + P(r = 1) +P(r = 2)
= 0.2824 + 0.3766 + 0.2301 = 0.8891

5. P r 2 = 1 - P(0) - P(1) = 1 - 0.2824 - 0.3766 = 0.3410

Example: In a large group of students 80% have a recommended statistics book. Three
students are selected at random. Find the probability distribution of the number of students
having the book. Also compute the mean and variance of the distribution.
Solution:
Let the event that ‘a student selected at random has the book’ be termed as a success. Since the
group of students is large, 3 trials, i.e., the selection of 3 students, can be regarded as independent
with probability of a success p = 0.8. Thus, the conditions of the given experiment satisfies the
conditions of binomial distribution.
3 r 3 r
The probability mass function P r C 0.8 0.2
r
where r = 0, 1, 2 and 3
The mean is np = 3 0.8 = 2.4 and Variance is npq = 2.4 0.2 = 0.48

Example:

1. The mean and variance of a discrete random variable X are 6 and 2 respectively. Assuming
X to be a binomial variate, find P(5 X 7) .

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