Question

Situation B: Suppose that 40% of Missourians favor increasing the minimum wage. Twelve Missourians are selected at random and asked whether they support the increase. Question B1: Find the probability that exactly three of the twelve support the increase.

Answer 1

Answer:

14.19% probability that exactly three of the twelve support the increase.

Step-by-step explanation:

For each Missourian, there are only two possible outcomes. Either they support the increase, or they do not. The probability of a Missourian supporting the increase is independent of other Missourians. 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.

40% of Missourians favor increasing the minimum wage.

This means that $p = 0.4$

Twelve Missourians are selected at random and asked whether they support the increase.

This means that $n = 12$

Find the probability that exactly three of the twelve support the increase.

This is P(X = 3).

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

$P(X = 3) = C_{12,3}.(0.4)^{3}.(0.6)^{9} = 0.1419$

14.19% probability that exactly three of the twelve support the increase.