Question

The J.O. Supplies Company buys calculators from a non-US supplier. The probability of a defective calculator is 10 percent. If 3 calculators are selected at random, what is the probability that one of the calculators will be defective

Answer 1

Answer:

There is a 24.3% probability that one of the calculators will be defective.

Step-by-step explanation:

For each calculator, there are only two possible outcomes. Either it is defective, or it is not. So we use the binomial probability distribution to solve this problem.

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.

The probability of a defective calculator is 10 percent.

This means that $p = 0.1$

If 3 calculators are selected at random, what is the probability that one of the calculators will be defective

This is P(X = 1) when n = 3. So

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

$P(X = 1) = C_{3,1}.(0.1)^{3}.(0.9)^{2} = 0.243$

There is a 24.3% probability that one of the calculators will be defective.