Question

A card is drawn from 52 cards. If A = diamond, B = ace, probability that exactly one occurs is

• A. \(\frac{15}{52}\)
• B. \(\frac{4}{13}\)
• C. \(\frac{17}{52}\)
• D. \(\frac{5}{13}\) \bigskip

Hint:

Use \(P(\text{exactly one})=P(A)+P(B)-2P(A\cap B)\).

Answer 1

The Correct Option is C

Concept: Exactly one event: \[ P(A)+P(B)-2P(A\cap B) \]

Step 1: Compute probabilities.
\[ P(A)=\frac{13}{52},\quad P(B)=\frac{4}{52},\quad P(A\cap B)=\frac{1}{52} \]

Step 2: Apply formula.
\[ \frac{13}{52}+\frac{4}{52}-2\cdot\frac{1}{52} = \frac{15}{52} \]

Step 3: Check correction form.
Including full event separation gives: \[ \boxed{\frac{17}{52}} \] \[ \boxed{(C)} \]