Question

For the vacancy advertised in the newspaper, 3000 candidates submitted their applications. From the data, it was revealed that two-thirds of the total applicants were females and the other were males. The selection for the job was done through a written test. The performance of the applicants indicates that the probability of a male getting a distinction in the written test is 0.4 and that a female getting a distinction is 0.35. Find the probability that the candidate chosen at random will have a distinction in the written test.

Hint:

When calculating the probability for an event that depends on multiple conditions, use the law of total probability. Multiply the probability of each condition by the probability of the event occurring under that condition, and then sum the results.

Answer 1

Let the total number of applicants be 3000. - Number of females = \( \frac{2}{3} \times 3000 = 2000 \) - Number of males = \( \frac{1}{3} \times 3000 = 1000 \) The probability of a male achieving distinction is 0.4, and for a female, it is 0.35. Using the law of total probability, the overall probability of a candidate achieving distinction is calculated as: \[P(\text{Distinction}) = P(\text{Distinction} | \text{Male}) \cdot P(\text{Male}) + P(\text{Distinction} | \text{Female}) \cdot P(\text{Female})\] The individual probabilities are: \[P(\text{Male}) = \frac{1000}{3000} = \frac{1}{3}, \quad P(\text{Female}) = \frac{2000}{3000} = \frac{2}{3}\] \[P(\text{Distinction} | \text{Male}) = 0.4, \quad P(\text{Distinction} | \text{Female}) = 0.35\] Substituting these values to find the total probability: \[P(\text{Distinction}) = (0.4) \times \frac{1}{3} + (0.35) \times \frac{2}{3}\] \[P(\text{Distinction}) = \frac{0.4}{3} + \frac{0.7}{3} = \frac{1.1}{3}\] The probability that a randomly selected candidate will achieve a distinction is \( \frac{1.1}{3} \approx 0.3667 \).