<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 5:</strong> \(P(E)\) is minimum when \(x\) equals to</p>
Step-by-Step Solution
Key Concept: The probability of getting sum x is the sum of probabilities from all possible outcomes (die A value, die B value) that add to x. We must calculate P(E) for each possible sum x ∈ {2,3,4,5,6} by considering the individual probability distributions of both dice.
<p><strong>Step 1: Determine individual probability distributions</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: Calculate P(E) for each possible sum x</strong></p><p>P(sum=2) = P(A=1)·P(B=1) = (1/2)(1/6) = 1/12</p><p>P(sum=3) = P(A=1)·P(B=2) + P(A=2)·P(B=1) = (1/2)(1/3) + (1/3)(1/6) = 1/6 + 1/18 = 4/18 = 2/9</p><p>P(sum=4) = P(A=1)·P(B=3) + P(A=2)·P(B=2) + P(A=3)·P(B=1) = (1/2)(1/2) + (1/3)(1/3) + (1/6)(1/6) = 1/4 + 1/9 + 1/36 = 9/36 + 4/36 + 1/36 = 14/36 = 7/18</p><p>P(sum=5) = P(A=2)·P(B=3) + P(A=3)·P(B=2) = (1/3)(1/2) + (1/6)(1/3) = 1/6 + 1/18 = 4/18 = 2/9</p><p>P(sum=6) = P(A=3)·P(B=3) = (1/6)(1/2) = 1/12</p><p><strong>Step 3: Compare probabilities</strong></p><p>Converting to common denominator 36: 3/36, 8/36, 14/36, 8/36, 3/36</p><p>Minimum value is 3/36 = 1/12, occurring at x = 2 and x = 6</p><p>∴ Answer: D (x = 2 or x = 6, depending on given options)</p>
Correct Answer: D