Question

The probability that A passes a test is \( \frac{2}{3} \) and the probability that B passes the same test is \( \frac{3}{5} \). The probability that only one passes is

• A. \( \frac{2}{5} \)
• B. \( \frac{4}{15} \)
• C. \( \frac{2}{15} \)
• D. \( \frac{7}{15} \)

Hint:

When dealing with probabilities of independent events, the probability of their intersection is the product of their individual probabilities. The probability of mutually exclusive events occurring is the sum of their individual probabilities.

Answer 1

The Correct Option is D

Probability that Only One Passes

Let \( P(A) \) be the probability that A passes the test, and \( P(B) \) be the probability that B passes the test. We are given:

\( P(A) = \frac{2}{3} \) 
\( P(B) = \frac{3}{5} \)

The probability that A fails the test is:

\( P(A') = 1 - P(A) = 1 - \frac{2}{3} = \frac{1}{3} \)

The probability that B fails the test is:

\( P(B') = 1 - P(B) = 1 - \frac{3}{5} = \frac{2}{5} \)

The event that only one passes occurs in two mutually exclusive ways:

1. A passes and B fails: \( A \cap B' \)
2. A fails and B passes: \( A' \cap B \)

Assuming the events are independent, we calculate:

\( P(A \cap B') = P(A) \times P(B') = \frac{2}{3} \times \frac{2}{5} = \frac{4}{15} \) 
\( P(A' \cap B) = P(A') \times P(B) = \frac{1}{3} \times \frac{3}{5} = \frac{3}{15} = \frac{1}{5} \)

The probability that only one passes is:

\( P(\text{only one passes}) = P(A \cap B') + P(A' \cap B) = \frac{4}{15} + \frac{3}{15} = \frac{7}{15} \)

Quick Tip: Use the principle of mutually exclusive events and independence to simplify such compound probability calculations.