Question

In a study, 39% of adults questioned reported that their health was excellent. A researcher wishes to study the health of people living close to a nuclear power plant. Among 10 adults randomly selected from this area, only 3 reported that their health was excellent. Find the probability that when 10 adults are randomly selected, 3 or fewer are in excellent health.

Answer 1

There is a 40.76% probability that when 10 adults are randomly selected, 3 or fewer are in excellent health.

Explanation:

For each adult questioned, there are only two possible outcomes. Either their health is excellent, or it is not. This means that we solve this problem using binomial probability concepts.

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).\pi^(x).(1-\pi)^(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) = (n!)/(x!(n-x)!)`

And
`\pi` is the probability of X happening.

In this problem

10 adults are asked, no
`n = 10`.

A success is an adult saying that their health was excellent. 39% of adults questioned reported that their health was excellent. This means that
`\pi = 0.39`.

Find the probability that when 10 adults are randomly selected, 3 or fewer are in excellent health.

This is
`P(X \leq 3)`.

`P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)`

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

`P(X = 0) = C_(10,0)*(0.39)^(0).(0.61)^(10) = 0.0071`

`P(X = 1) = C_(10,1)*(0.39)^(1).(0.61)^(9) = 0.0456`

`P(X = 2) = C_(10,2)*(0.39)^(2).(0.61)^(8) = 0.1312`

`P(X = 3) = C_(10,3)*(0.39)^(3).(0.61)^(7) = 0.2237`

Finally

`P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0071 + 0.0456 + 0.1312 + 0.2237 = 0.4076`

There is a 40.76% probability that when 10 adults are randomly selected, 3 or fewer are in excellent health.