Question

Three students \( X, Y, Z \) appear for an exam. Their passing probabilities are: \[ P(X) = \frac{1}{5},\quad P(Y) = \frac{1}{4},\quad P(Z) = 1 - \frac{2}{3} = \frac{1}{3} \] What is the probability that at least two of them pass?

• A. \( \frac{1}{6} \)
• B. \( \frac{2}{5} \)
• C. \( \frac{3}{4} \)
• D. \( \frac{3}{5} \)

Hint:

When calculating "at least" probabilities, subtract complementary cases from 1. Consider none, one, and two successes.

Answer 1

The Correct Option is A

We need \( P(\text{at least 2 pass}) = 1 - P(\text{none pass}) - P(\text{only one passes}) \) Let: \[ P(X') = \frac{4}{5},\quad P(Y') = \frac{3}{4},\quad P(Z') = \frac{2}{3} \] 1. None pass: \[ P(X')P(Y')P(Z') = \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{2}{3} = 1.0 \Rightarrow \text{(Contradiction—check again)} \] Actually compute: None pass: \[ P(X' \cap Y' \cap Z') = \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{2}{3} = \frac{24}{60} = \frac{2}{5} \] Only one passes: - X passes: \( \frac{1}{5} \cdot \frac{3}{4} \cdot \frac{2}{3} = \frac{6}{60} \) - Y passes: \( \frac{4}{5} \cdot \frac{1}{4} \cdot \frac{2}{3} = \frac{8}{60} \) - Z passes: \( \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{1}{3} = \frac{12}{60} \) Total = \( \frac{6 + 8 + 12}{60} = \frac{26}{60} = \frac{13}{30} \) Now: \[ P(\text{at least 2 pass}) = 1 - \left( \frac{2}{5} + \frac{13}{30} \right) = 1 - \left( \frac{12 + 13}{30} \right) = 1 - \frac{25}{30} = \frac{1}{6} \]