Suppose a recent government report indicates that 12% of the labor force is Afri

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.

Explanation / Answer

Back-up Theory

If A and B are two events such that probability of B is influenced by occurrence or otherwise of A, then

Conditional Probability of B given A, denoted by P(B/A) = P(B A)/P(A)..….(1)

P(B) = {P(B/A) x P(A)} + {P(B/AC) x P(AC)}………………………………….(2)

P(A/B) = P(B/A) x { P(A)/P(B)}……………………………..………………….(3)

With events A and B as defined in the question, given,

‘12% of labor force is is AA (African-American)’ => P(A) = 0.12 and hence

P(AC) = 1 – 0.12 = 0.88.

‘of individuals in the labor force who are AA, 11% are unemployed’ => P(B/A) = 0.11.

‘of individuals in the labor force who are not AA, 5% are unemployed’ => P(B/AC) = 0.05.

Now, [vide (2) under Back-up Theory], P(B) = {(0.11 x 0.12) + (0.05 x 0.88)} = 0.0572.

Also, [vide (3) under Back-up Theory], P(A/B) = 0.11 x { 0.12/0.0572} = 0.0132/0.0572

= 0.2308 = 23% ANSWER

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