Question

A study conducted at a certain college shows that 54% of the school's graduates move to a different state after graduating. Find the probability that among 7 randomly selected graduates, at least one moves to a different state after graduating.

Answer 1

99.56% probability that among 7 randomly selected graduates, at least one moves to a different state after graduating.

Explanation:

For each graduate, there are only two possible outcomes. Either they move to a different state, or they do not. The probability of a graduate moving to a different state is independent of other graduates. 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.

54% of the school's graduates move to a different state after graduating.

This means that
`p = 0.54`

7 randomly selected graduates

This means that
`n = 7`

Find the probability that among 7 randomly selected graduates, at least one moves to a different state after graduating.

Either none moves, or at least one does. The sum of the probabilities of these events is 1. So

`P(X = 0) + P(X \geq 1) = 1`

We want
`P(X \geq 1)`. Then

`P(X \geq 1) = 1 - P(X = 0)`

In which

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

`P(X = 0) = C_(7,0).(0.54)^(0).(0.46)^(7) = 0.0044`

`P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0044 = 0.9956`

99.56% probability that among 7 randomly selected graduates, at least one moves to a different state after graduating.