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.
Required:
What is the probability that the lot will pass inspection?

Answer 1

Answer:

0.0016 = 0.16% probability that the lot will pass inspection.

Step-by-step explanation:

For each bulb, there are only two possible outcomes. Either it is defective, or it is not. The probability of a bulb being defective is independent of any other bulb, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Sample of 18 light bulbs

This means that $n = 18$

30% of the bulbs in the lot are defective.

This means that $p = 0.3$

What is the probability that the lot will pass inspection?

It will pass inspection if there are no defective bulbs, that is, we have to find P(X = 0). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{18,0}.(0.3)^{0}.(0.7)^{18} = 0.0016$

0.0016 = 0.16% probability that the lot will pass inspection.