Question

At a city council meeting, there 101 people present, and 66 were elderly.

If you talked to six random people there, what is the percent chance that exactly three is elderly? (Round
to the nearest hundredth of a percent.)

Answer 1

23.63% chance that exactly three is elderly

Explanation:

The people are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

`P(X = x) = h(x,N,n,k) = (C_(k,x)*C_(N-k,n-x))/(C_(N,n))`

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

`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)!)`

In this question:

101 total people means that
`N = 101`

Sample of 6 means that
`n = 6`

66 elderly means that
`k = 66`

If you talked to six random people there, what is the percent chance that exactly three is elderly?

This is P(X = 3). So

`P(X = x) = h(x,N,n,k) = (C_(k,x)*C_(N-k,n-x))/(C_(N,n))`

`P(X = 3) = h(3,101,6,66) = (C_(66,3)*C_(35,3))/(C_(101,6)) = 0.2363`

0.2363*100% = 23.63%

23.63% chance that exactly three is elderly