1. The problem states: There are 100 applicants with 56 males (m) and 44 females (f). Given probabilities: - Probability of acceptance given male applicant $p(a/m) = \frac{1}{5}$ - Probability of acceptance given female applicant $p(a/f) = \frac{1}{11}$ We are asked to find the probability that the applicant is male given acceptance $p(m/a)$ and the probability that the applicant is female given acceptance $p(f/a)$. 2. Use Bayes' theorem which states: $$ p(m/a) = \frac{p(a/m) \times p(m)}{p(a)}\quad\text{and}\quad p(f/a) = \frac{p(a/f) \times p(f)}{p(a)} $$ 3. Calculate $p(m)$ and $p(f)$, the base probabilities of males and females: - $p(m) = \frac{56}{100} = 0.56$ - $p(f) = \frac{44}{100} = 0.44$ 4. Calculate $p(a)$, the total acceptance probability using the law of total probability: $$ p(a) = p(a/m) \times p(m) + p(a/f) \times p(f) = \frac{1}{5} \times 0.56 + \frac{1}{11} \times 0.44 $$ Calculate each term: - $\frac{1}{5} \times 0.56 = 0.112$ - $\frac{1}{11} \times 0.44 \approx 0.04$ Therefore, $$ p(a) = 0.112 + 0.04 = 0.152 $$ 5. Calculate $p(m/a)$: $$ p(m/a) = \frac{p(a/m) \times p(m)}{p(a)} = \frac{0.112}{0.152} \approx 0.7368 $$ 6. Calculate $p(f/a)$: $$ p(f/a) = \frac{p(a/f) \times p(f)}{p(a)} = \frac{0.04}{0.152} \approx 0.2632 $$ 7. Final answers: - Probability the applicant is male given acceptance $p(m/a) \approx 0.7368$ - Probability the applicant is female given acceptance $p(f/a) \approx 0.2632$