In a certain company, 40% of the employees are females. Suppose 60% of the male
workers are married and 40% of the female workers are married. What is the
probability that a married worker is male?
My proof:
$$A = {Female}$$
$$A^c = Male $$
$$B = Married$$
So, $$P(A^c) = 0.6$$
$$P(B/A^c) = 0.6 * 0.6 = 0.36$$
I believe this means the probability of randomly picking a married male. So, I do not know if this is the answer or I keep going:
$$P(A) = 0.4$$
$$P(B/A) = 0.4 * 0.4 = 0.16$$
So, $$P((B/A^c) ∪ (B/A)) = ((B/A^c) + (B/A)) -((B/A^c)(B/A)) $$
$$(0.52) – (0.0576) = 0.4626 $$=PROBABILITY OF PICKING A RANDOM MARRIED EMPLOYEE AT RANDOM
So, $$P(A^c ∩ ((B/A^c) ∪ (B/A))) = 0.6 * 0.4626 = .2744$$
Which is my final answer, but I am very confused by this question since it seems like a small probability and apparently 60% of workers are male and 60% of those male workers are married
Best Answer
Suppose there are exactly $100$ employees at this company. If $40\%$ of workers are female, then $40$ are female and $100 - 40 = 60$ are male. If $40\%$ of these female workers are married, then there are $40(0.4) = 16$ married female workers and $40 - 16 = 24$ unmarried female workers. If $60\%$ of the $60$ male workers are married, then there are $60(0.6) = 36$ married male workers, and $60 - 36 = 24$ unmarried male workers. We summarize this in the following table: $$\begin{array}{c|c|c|c} & \text{Female} & \text{Male} & \text{Total} \\ \hline \text{Married} & 16 & 36 & 52 \\ \hline \text{Unmarried} & 24 & 24 & 48 \\ \hline \text{Total} & 40 & 60 & 100 \end{array}$$ Therefore, among the $16 + 36 = 52$ married workers of either sex, $36$ are males, and if one married worker is chosen at random, the chance this person is male is simply $36/52 = 9/13 \approx 0.692308$.