Question

Based on the U.S. Census Bureau’s American Community Survey of 2017, 12.9% of the U.S. population was foreign-born. The U.S. Census Bureau uses the term foreign-born to refer to anyone who is not a U.S. citizen at birth. When I took a sample of 5 students from my statistics class find the following probabilities:

1 Find the probability that none of the students are foreign-born (x=0)
2 find p(x>=1) (that at least one is foreign-born)

Answer 1

1) 50.13% probability that none of the students are foreign-born (x=0)

2) 49.87% probability that at least one is foreign-born.

Explanation:

For each student sampled, there are only two possible outcomes. Either they are foreign-born, or they are not. This means that we use the binomial probability distribution to solve this problem.

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 combinatios 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 this problem we have that:

`n = 5, p = 0.129`

1 Find the probability that none of the students are foreign-born (x=0)

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

`P(X = 0) = C_(5,0).(0.129)^(0).(0.871)^(5) = 0.5013`

There is a 50.13% probability that none of the students are foreign-born (x=0)

2 find p(x>=1) (that at least one is foreign-born)

Either no students are foreign-born, or at least one is. The sum of the probabilities of these events is decimal 1.

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

`P(X \geq 1) = 1 - P(X = 0) = 1 - 0.5013 = 0.4987`

There is a 49.87% probability that at least one is foreign-born.