Question

Find the probability that exactly one of them is selected.

Hint:

To calculate "exactly one" selection, consider all cases where one succeeds, and the others fail, then sum the probabilities.

Answer 1

Step 1: Compute the probability of exactly one being selected
The probability of exactly one being selected is: \[ P(\text{Exactly one selected}) = P(R) \cdot P(\overline{J}) \cdot P(\overline{A}) + P(\overline{R}) \cdot P(J) \cdot P(\overline{A}) + P(\overline{R}) \cdot P(\overline{J}) \cdot P(A), \] where: \[ P(\overline{J}) = 1 - P(J) = \frac{2}{3}, \quad P(\overline{A}) = 1 - P(A) = \frac{3}{4}, \quad P(\overline{R}) = 1 - P(R) = \frac{4}{5}. \] Substitute the values: \[ P(\text{Exactly one selected}) = \frac{1}{5} \cdot \frac{2}{3} \cdot \frac{3}{4} + \frac{4}{5} \cdot \frac{1}{3} \cdot \frac{3}{4} + \frac{4}{5} \cdot \frac{2}{3} \cdot \frac{1}{4}. \] Simplify each term: \[ P(\text{Exactly one selected}) = \frac{6}{60} + \frac{12}{60} + \frac{8}{60} = \frac{26}{60} = \frac{13}{30}. \] Final Result: The probability that exactly one of them is selected is \( \frac{13}{30} \).