Question

If $ A $ and $ B $ are two events such that $ P(A) = 0.7 $, $ P(B) = 0.4 $ and $ P\left( A \cap \overline{B} \right) = 0.5 $, where $\overline{B}$ denotes the complement of $ B $, then $ P\left( B | \left( A \cup \overline{B} \right) \right) $ is equal to

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

Hint:

- Remember \( P(A \cap \overline{B}) = P(A) - P(A \cap B) \) - \( A \cup \overline{B} \) can be visualized using Venn diagrams - For conditional probability \( P(X|Y) \), both numerator and denominator must relate to the same probability space - Simplify complex events using probability identities

Answer 1

The Correct Option is B

Step 1: Find \( P(A \cap B) \)
\[ P(A \cap \overline{B}) = P(A) - P(A \cap B) \] \[ 0.5 = 0.7 - P(A \cap B) \] \[ P(A \cap B) = 0.2 \]

Step 2: Calculate \( P(A \cup \overline{B}) \)
Using probability rules: \[ P(A \cup \overline{B}) = P(A) + P(\overline{B}) - P(A \cap \overline{B}) \] \[ = 0.7 + (1 - 0.4) - 0.5 \] \[ = 0.7 + 0.6 - 0.5 = 0.8 \]

Step 3: Find \( P(B \cap (A \cup \overline{B})) \)
\[ P(B \cap (A \cup \overline{B})) = P((B \cap A) \cup (B \cap \overline{B})) \] \[ = P(A \cap B) + P(\emptyset) = 0.2 + 0 = 0.2 \]

Step 4: Compute conditional probability
\[ P(B | A \cup \overline{B}) = \frac{P(B \cap (A \cup \overline{B}))}{P(A \cup \overline{B})} = \frac{0.2}{0.8} = \frac{1}{4} \]