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