Question

Statistics show that about 42% of Americans voted in the previous national election. If three Americans are randomly selected, what is the probability that none of them voted in the last election

Answer 1

Answer:

19.51% probability that none of them voted in the last election

Step-by-step explanation:

For each American, there are only two possible outcomes. Either they voted in the previous national election, or they did not. The probability of an American voting in the previous election is independent of other Americans. 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.

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

And p is the probability of X happening.

42% of Americans voted in the previous national election.

This means that $p = 0.42$

Three Americans are randomly selected

This means that $n = 3$

What is the probability that none of them voted in the last election

This is P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{3,0}.(0.42)^{0}.(0.58)^{3} = 0.1951$

19.51% probability that none of them voted in the last election