Question

Suppose a recent government report indicates that 12% of the labor force is African American. Of the individuals in the labor force who arc African American, 11% are unemployed. Among the individuals in the labor force who arc 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:

P(A|B) = 0.231 = 23%

Step-by-step explanation:

A is the event that a randomly selected member of the Labor force is African American, P(A) = 0.12 (given in the question)

Non African-Americans in the labour force, P(A') = 1 - P(A') = 1 - 0.12 = 0.88

B is the event that a member of the labor force is unemployed, P(B) is not directly given, but can be calculated from the given percentages on the question.

Total unemployed people in the labour force = unemployed African American people in the labor force + unemployed people in the labor force that aren't African Americans

African-Americans in the labour force P(A) = 0.12

Non African-Americans in the labour force, P(A') = 0.88

unemployed African Americans in the labor force P(A n B) = 0.11 of the African Americans in the labor force = 0.12 × 0.11 = 0.0132

Unemployed Non-African Americans in the labor force, P(A' n B) = 5% of 0.88 = 0.88 × 0.05 = 0.044

Total unemployed Americans in the labor force, P(B) = P(A n B) + P(A' n B) = 0.0132 + 0.044 = 0.0572

Probability that a person is African American, given that they're unemployed, P(A|B) = (P(A n B))/P(B) = 0.0132/0.0572 = 0.231 = 23%