Sixty percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive vehicles pass or fail independently of one another, calculate the following probabilities. (Enter your answers to three decimal places.)

(a) P(all of the next three vehicles inspected pass)
(b) P(at least one of the next three inspected fails)

For each vehicle examined there are only two possible outcomes. Either they pass the inspection, or they do not. The probabilities for each vehicle are independent. 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.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

In this problem we have that:

Sixty percent of all vehicles examined at a certain emissions inspection station pass the inspection. This means that [tex]p = 0.6[/tex]

(a) P(all of the next three vehicles inspected pass)

This is [tex]P(X = 3)[/tex] when [tex]n = 3[/tex]. So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 3) = C_{3,3}.(0.6)^{3}.(0.4)^{0} = 0.216[/tex]

(b) P(at least one of the next three inspected fails)

Either all three vehicles inspected pass, or at least one fails. The sum of the probabilities of these events is decimal 1. Mathematically,

[tex]P(X = 3) + P(X < 3) = 1[/tex]

We want [tex]P(X < 3)[/tex]

So

[tex]P(X < 3) = 1 - P(X = 3)[/tex]

From a), [tex]P(X = 3) = 0.216[/tex]. So

[tex]P(X < 3) = 1 - P(X = 3) = 1 - 0.216 = 0.784[/tex]