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

Step-by-step 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)

$=^{15}C_3(0.20)^{3}(1-0.20)^{12}\\\\=\dfrac{15!}{3!12!}(0.20)^3(0.80)^{12}=0.2501$

Required probability = 0.2501