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

Answer:

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.

Step-by-step 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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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.