Question

Approximately 5% of calculators coming out of the production lines have a defect. Fifty calculators are randomly selected from the production line and tested for defects. What is the probability that exactly 2 calculators are defective?

Answer 1

Answer:

0.2611 = 26.11% probability that exactly 2 calculators are defective.

Step-by-step explanation:

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

5% of calculators coming out of the production lines have a defect.

This means that $p = 0.05$

Fifty calculators are randomly selected from the production line and tested for defects.

This means that $n = 50$

What is the probability that exactly 2 calculators are defective?

This is P(X = 2). So

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

$P(X = 2) = C_{50,2}.(0.05)^{2}.(0.95)^{48} = 0.2611$

0.2611 = 26.11% probability that exactly 2 calculators are defective.