Question

A production process produces 2% defective parts. A sample of nine parts from the production process is selected. What is the probability that the sample contains exactly two defective parts? a. .0125 b. .9875 c. .9994 d. .0200

Answer 1

Answer:

a. .0125

Step-by-step explanation:

For each part, there are only two possible outcomes. Either it is defective, or it is not. The probability of a part being defective is independent of any other part. This 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.

2% defective parts.

This means that $p = 0.02$

A sample of nine parts from the production process is selected.

This means that $n = 9$

What is the probability that the sample contains exactly two defective parts?

This is P(X = 2).

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

$P(X = 2) = C_{9,2}.(0.02)^{2}.(0.98)^{7} = 0.0125$

0.0125 = 1.25% probability that the sample contains exactly two defective parts. The correct answer is given by option a.