Question

\(P(A) = \frac{1}{3}\), \(P(B) = \frac{3}{4}, P(A \cap B) = \frac{1}{6}\), then probability of \( A \) alone is

• A. \( \frac{1}{3} \)
• B. \( \frac{1}{2} \)
• C. \( \frac{1}{6} \)
• D. \( \frac{1}{8} \)

Hint:

The probability of event \( A \) alone is \( P(A \cap B^c) = P(A) - P(A \cap B) \). This represents the probability of \( A \) occurring without \( B \).

Answer 1

The Correct Option is C

Step 1: Interpret "probability of \( A \) alone."
The probability of \( A \) alone typically means the probability of event \( A \) occurring but not \( B \), i.e., \( P(A \cap B^c) \), where \( B^c \) is the complement of \( B \). Step 2: Compute \( P(A \cap B^c) \).
We are given:
\( P(A) = \frac{1}{3} \),
\( P(B) = \frac{3}{4} \),
\( P(A \cap B) = \frac{1}{6} \).
First, find \( P(B^c) \): \[ P(B^c) = 1 - P(B) = 1 - \frac{3}{4} = \frac{1}{4}. \] Use the formula for \( P(A \cap B^c) \): \[ P(A \cap B^c) = P(A) - P(A \cap B). \] Substitute the given values: \[ P(A \cap B^c) = \frac{1}{3} - \frac{1}{6} = \frac{2}{6} - \frac{1}{6} = \frac{1}{6}. \] So, the probability of \( A \) alone is \( \frac{1}{6} \). Step 3: Verify independence (optional, for context).
Check if \( A \) and \( B \) are independent: \[ P(A) \cdot P(B) = \frac{1}{3} \cdot \frac{3}{4} = \frac{3}{12} = \frac{1}{4}, \] but \( P(A \cap B) = \frac{1}{6} \neq \frac{1}{4} \), so \( A \) and \( B \) are not independent. This does not affect the calculation of \( P(A \cap B^c) \). Step 4: Evaluate the options.
(1) \( \frac{1}{3} \): Incorrect, as \( P(A \cap B^c) \neq P(A) \). Incorrect.
(2) \( \frac{1}{2} \): Incorrect, as the value is too large. Incorrect.
(3) \( \frac{1}{6} \): Correct, matches the computed value. Correct.
(4) \( \frac{1}{8} \): Incorrect, as the value is too small. Incorrect.
Step 5: Select the correct answer.
The probability of \( A \) alone is \( \frac{1}{6} \), matching option (3).