Question

A quality control inspector has drawn a sample of 15 light bulbs from a recent production lot. Suppose 20% of the bulbs in the lot are defective. What is the probability that exactly 3 bulbs from the sample are defective

Answer 1

Answer: 0.2501

Explanation:

Given : The proportion of the bulbs in the lot are defective = 20%= 0.20

Sample size : n =15

Let x be the binomial variable that represents the defective bulbs.

Binomial probability formula :
`^nC_xp^x(1-p)^x`

The probability that exactly 3 bulbs from the sample are defective : P(X=3)

`=^(15)C_3(0.20)^(3)(1-0.20)^(12)\\\\=(15!)/(3!12!)(0.20)^3(0.80)^(12)=0.2501`

Required probability = 0.2501