Question

Five percent of all blue ray players manufactured by a large electronics company are defective. A quality control inspector randomly selects three players from the production line. What is the probability that exactly one of these three blue ray players is defective

Answer 1

Answer:

0.1354 = 13.54% probability that exactly one of these three blue ray players is defective.

Step-by-step explanation:

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

Five percent of all blue ray players manufactured by a large electronics company are defective.

This means that $p = 0.05$

A quality control inspector randomly selects three players from the production line.

This means that $n = 3$

What is the probability that exactly one of these three blue ray players is defective?

This is P(X = 1). So

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

$P(X = 1) = C_{3,1}.(0.05)^{1}.(0.95)^{2} = 0.1354$

0.1354 = 13.54% probability that exactly one of these three blue ray players is defective.