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 1 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

0.9983

Explanation:

Given that :

Number of samples = 18

P(defective bulb) = 30% = 0.3

Inspection failure = defective bulb is 1 or more

Hence, inspection failure occurs if;

P(x ≥ 1) = 1 - P(x < 1)

P(x ≥ 1) = 1 - P(x = 0)

Using the binomial probability function :

nCx * p^x * (1 - p)^(n-x)

p = probability of success per trial ; n = number of trials

n = 18

P(x = 0) = [ 1 - (18C0 * 0.3^0 * (1 - 0.3)^(18-0))

= [1 - (1 * 1 * 0.7^18)]

1 - 0.0016284136

= 0.9983