<p><strong>For Problems 4–6</strong><br>There are two die \(A\) and \(B\) both having six faces. Die \(A\) has three faces marked with 1, two faces marked with 2, and one face marked with 3. Die \(B\) has one face marked with 1, two faces marked with 2, and three faces marked with 3. Both dices are thrown randomly once. If \(E\) be the event of getting sum of the numbers appearing on top faces equal to \(x\) and let \(P(E)\) be the probability of event \(E\), then</p><p><strong>Problem 6:</strong> When \(x = 4\), then \(P(E)\) is equal to</p>
Step-by-Step Solution
Key Concept: Sum equals 4 requires identifying all pairs (a,b) where a + b = 4, then calculating probability using the individual face frequencies: P(Die A shows i) × P(Die B shows j) for each valid pair, then summing all probabilities.
<p><strong>Step 1: Identify the probability distribution for each die</strong></p><p>Die A: P(1) = 3/6 = 1/2, P(2) = 2/6 = 1/3, P(3) = 1/6</p><p>Die B: P(1) = 1/6, P(2) = 2/6 = 1/3, P(3) = 3/6 = 1/2</p><p><strong>Step 2: Find all pairs (a,b) where a + b = 4</strong></p><p>The possible pairs are: (1,3), (2,2), (3,1)</p><p><strong>Step 3: Calculate probability for each pair</strong></p><p>P(A=1, B=3) = P(A=1) × P(B=3) = (1/2) × (1/2) = 1/4</p><p>P(A=2, B=2) = P(A=2) × P(B=2) = (1/3) × (1/3) = 1/9</p><p>P(A=3, B=1) = P(A=3) × P(B=1) = (1/6) × (1/6) = 1/36</p><p><strong>Step 4: Sum all probabilities</strong></p><p>P(E) = 1/4 + 1/9 + 1/36</p><p>Finding common denominator (36): P(E) = 9/36 + 4/36 + 1/36 = 14/36 = 7/18</p><p>∴ Answer: C (which equals 7/18)</p>
Correct Answer: C