Question

Three students A, B and C write an entrance exam. The chance that at least one of them passes the exam is \( \frac{3}{4} \). If the probability that A and C pass the exam are respectively \( \frac{1}{2} \) and \( \frac{1}{4} \), then the probability that B does not pass that exam is:

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

Hint:

Use the complement rule: \( P(\text{At least one}) = 1 - P(\text{None}) \). Then plug known values and solve.

Answer 1

The Correct Option is B

Let \( P(A') \), \( P(B') \), and \( P(C') \) denote the probabilities of A, B, and C failing, respectively. We are given: \[ P(\text{At least one passes}) = 1 - P(A' \cap B' \cap C') = \frac{3}{4} \Rightarrow P(A' \cap B' \cap C') = \frac{1}{4} \] Given: \[ P(A) = \frac{1}{2} \Rightarrow P(A') = \frac{1}{2}, \quad P(C) = \frac{1}{4} \Rightarrow P(C') = \frac{3}{4} \] So, \[ P(A') \cdot P(B') \cdot P(C') = \frac{1}{4} \Rightarrow \frac{1}{2} \cdot P(B') \cdot \frac{3}{4} = \frac{1}{4} \Rightarrow P(B') = \frac{1}{4} \cdot \frac{2}{1} \cdot \frac{4}{3} = \frac{2}{3} \]