Question

Suppose a recent government report indicates that 6% of the labor force is Asian. Of the individuals in the labor force who are Asian, 5% are unemployed. Among the individuals in the labor force who are not Asian, 6% are unemployed. Let be the event that a randomly selected member of the labor force is Asian and let be the event that a randomly selected member of the labor force is unemployed. Determine (∣) , the probability that a randomly selected member of the labor force is Asian given that he or she is unemployed. Express your answer as a percentage with no decimal places.

Answer 1

The probability that a randomly selected member of the labor force is Asian given that he or she is unemployed is 4.98%

Explanation:

We want to find the probability that a randomly selected member of the labor force is Asian given that he or she is unemployed, which can be written as P(A|U). Using Bayes' theorem, we have:

P(A|U) = P(U|A) * P(A) / P(U)

We are given P(A) = 0.06 (6% of the labor force is Asian), P(U|A) = 0.05 (among Asians in the labor force, 5% are unemployed), and P(U|not A) = 0.06 (among non-Asians in the labor force, 6% are unemployed).

To find P(U), we can use the law of total probability:

P(U) = P(U|A) * P(A) + P(U|not A) * P(not A)

We know P(A) = 0.06, so P(not A) = 1 - P(A) = 0.94.

Therefore,

P(U) = 0.05 * 0.06 + 0.06 * 0.94 = 0.0602

Now we can substitute all the values into Bayes' theorem:

P(A|U) = 0.05 * 0.06 / 0.0602 = 0.0498

So the probability that a randomly selected member of the labor force is Asian given that he or she is unemployed is 4.98% (rounded to the nearest percentage with no decimal places).