Question

An inspector examines a large shipment of several thousand laser pointers by randomly selecting and then inspecting 6 laser pointers. Suppose that 1% of the laser points are defective. Which one of the following expressions describes the probability that exactly one of the six laser pointers is defective?

Answer 1

Answer:

$P(X = 1) = C_{6,1}.(0.01)^{1}.(0.99)^{5} = 0.0571$

Step-by-step explanation:

For each laser pointer, there are only two possible outcomes. Eithey they are defective, or they are not. This means that 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 combinatios 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.

In this problem we have that:

$p = 0.01, n = 6$

Which one of the following expressions describes the probability that exactly one of the six laser pointers is defective?

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

$P(X = 1) = C_{6,1}.(0.01)^{1}.(0.99)^{5} = 0.0571$