Question

A quality control inspector has drawn a sample of 18 light bulbs from a recent production lot. If the number of defective bulbs is 2 or more, the lot fails inspection. Suppose 30% of the bulbs in the lot are defective. What is the probability that the lot will fail inspection? Round your answer to four decimal places.

Answer 1

Answer:

Given,

To determine the probability that the lot will fail inspection

let us consider,

p (Fail) = 30% = 0.30

n = 18

x = 2

Now we have to determine the required probability i.e., given below

= P(X >= 2)

= 1 - P(X < 2)

= 1- ( P(X = 0,1))

= 1- ( 18C0 *0.30^0 *0.70^18 + 18C1 *0.30^1 *0.70^17)

= 1 - 0.1419

P(X >= 2) = 0.9858