Question

A machine that manufactures automobile parts produces defective parts 16% of the time. If 9 parts produced by this machine are randomly selected, what is
the probability that fewer than 3 of the parts are defective?
Carry your intermediate computations to at least four decimal places, and round your answer to two decimal places.

Answer 1

Answer:

0.84 = 84% probability that fewer than 3 of the parts are defective.

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, and thus, 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.

A machine that manufactures automobile parts produces defective parts 16% of the time.

This means that $p = 0.16$

9 parts:

This means that $n = 9$

What is the probability that fewer than 3 of the parts are defective?

This is:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

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

$P(X = 0) = C_{9,0}.(0.16)^{0}.(0.84)^{9} = 0.2082$

$P(X = 1) = C_{9,1}.(0.16)^{1}.(0.84)^{8} = 0.3569$

$P(X = 2) = C_{9,2}.(0.16)^{2}.(0.84)^{7} = 0.2720$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2082 + 0.3569 + 0.2720 = 0.8371$

Rounding to two decimal places, 0.84.

0.84 = 84% probability that fewer than 3 of the parts are defective.