Question

Let \(E_1, E_2, E_3\) and \(E_4\) be four events such that \[ P(E_i|E_4) = \frac{2}{3}, \; i = 1, 2, 3; \quad P(E_i \cap E_j^c | E_4) = \frac{1}{6}, \; i,j = 1,2,3; \; i \ne j; \quad P(E_1 \cap E_2 \cap E_3^c | E_4) = \frac{1}{6}. \] Then, \( P(E_1 \cup E_2 \cup E_3 | E_4) \) is equal to

• A. \(\frac{1}{2}\)
• B. \(\frac{2}{3}\)
• C. \(\frac{5}{6}\)
• D. \(\frac{7}{12}\)

Hint:

When multiple event probabilities are conditioned on another event, inclusion–exclusion remains valid in conditional form — always compute pairwise and triple intersections carefully.

Answer 1

The Correct Option is C

Step 1: Use inclusion–exclusion principle.
We have \[ P(E_1 \cup E_2 \cup E_3 | E_4) = \sum_{i=1}^{3} P(E_i|E_4) - \sum_{i<j} P(E_i \cap E_j|E_4) + P(E_1 \cap E_2 \cap E_3|E_4). \]
Step 2: Substitute given values.
Each \( P(E_i|E_4) = \frac{2}{3} \), so \[ \sum P(E_i|E_4) = 3 \times \frac{2}{3} = 2. \] Also, we are given \( P(E_i \cap E_j^c|E_4) = \frac{1}{6} \). Using the identity \[ P(E_i|E_4) = P(E_i \cap E_j|E_4) + P(E_i \cap E_j^c|E_4), \] we get \[ \frac{2}{3} = P(E_i \cap E_j|E_4) + \frac{1}{6} \Rightarrow P(E_i \cap E_j|E_4) = \frac{1}{2}. \] Hence, \[ \sum_{i<j} P(E_i \cap E_j|E_4) = 3 \times \frac{1}{2} = \frac{3}{2}. \]
Step 3: Find \( P(E_1 \cap E_2 \cap E_3|E_4) \).
We are given \( P(E_1 \cap E_2 \cap E_3^c|E_4) = \frac{1}{6} \). Thus, \[ P(E_1 \cap E_2|E_4) = P(E_1 \cap E_2 \cap E_3|E_4) + P(E_1 \cap E_2 \cap E_3^c|E_4). \] \[ \frac{1}{2} = P(E_1 \cap E_2 \cap E_3|E_4) + \frac{1}{6} \Rightarrow P(E_1 \cap E_2 \cap E_3|E_4) = \frac{1}{3}. \]
Step 4: Apply inclusion–exclusion.
\[ P(E_1 \cup E_2 \cup E_3 | E_4) = 2 - \frac{3}{2} + \frac{1}{3} = \frac{5}{6}. \] Final Answer: \[ \boxed{\frac{5}{6}} \]