Question

On a particular day, 112 of 280 passengers on a particular DTW-LAX flight used the e-ticket check-in kiosk to obtain boarding passes. In a random sample of eight passengers, use the binomial model to find the approximate hypergeometric probability that four will have used the e-ticket check-in kiosk to obtain boarding passes.

Answer 1

0.2348 = 23.48% probability that four will have used the e-ticket check-in kiosk to obtain boarding passes.

Explanation:

Hypergeometric distribution:

The probability of x sucesses 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 sucesses.

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:

Population of 280, which means that
`N = 280`

112 use e-ticket check-in kiosk, which means that
`k = 112`

Sample of eight passengers means that
`n = 8`

Find the approximate hypergeometric probability that four will have used the e-ticket check-in kiosk to obtain boarding passes.

This is P(X = 4). So

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

`P(X = 4) = h(4,280,8,112) = (C_(112,4)*C_(168,4))/(C_(280,4)) = 0.2348`

0.2348 = 23.48% probability that four will have used the e-ticket check-in kiosk to obtain boarding passes.