Question

In one town, 79% of adults have health insurance. What is the probability that 4 adults selected at random from the town all have health insurance? Consider an independent events and round to the nearest thousandth if necessary g

Answer 1

0.39 = 39% probability that 4 adults selected at random from the town all have health insurance

Explanation:

For each adult, there are only two possible outcomes. Either they have health insurance, or they do not. The probability of an adult having health insurance is independent of other adults. So we use the binomial probability distribution 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) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

In one town, 79% of adults have health insurance

This means that
`p = 0.79`

4 adults

This means that
`n = 4`

What is the probability that 4 adults selected at random from the town all have health insurance?

This is P(X = 4).

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 4) = C_(4,4).(0.79)^(4).(0.21)^(0) = 0.39`

0.39 = 39% probability that 4 adults selected at random from the town all have health insurance