Question

Suppose a recent government report indicates that 12% of the labor force is African American. Of the individuals in the labor force who are African American, 11% are unemployed. Among the individuals in the labor force who are not African American, 5% are unemployed. Let A be the event that a randomly selected member of the labor force is African American and let B be the event that a randomly selected member of the labor force is unemployed. Determine P ( A ∣ B ) , the probability that a randomly selected member of the labor force is African American given that he or she is unemployed. Express your answer as a percentage with no decimal places.

Answer 1

Answer:

The probability that a randomly selected member of the labor force is African American given that he or she is unemployed is 0.2308.

Explanation:

The events are denoted as:

A = a member of a labor force is African American

B =  a member of a labor force is unemployed

The information provided is:

$P(A)=0.12\\P(B|A)=0.11\\P(B|A^{c})=0.05$

The Bayes' theorem states that the conditional probability of an event E given that another event X has already occurred is:

$P(E|X)=\frac{P(X|E)P(E)}{P(X|E)P(E)+P(X|E^{c})P(E^{c})}$

Use the Bayes' theorem to compute the value of P (A|B) as follows:

$P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^{c})P(A^{c})}=\frac{0.11\times 0.12}{(0.11\times 0.12)+(0.05\times (1-0.12))}=0.2308$

Thus, the probability that a randomly selected member of the labor force is African American given that he or she is unemployed is 0.2308.