Probability
Classical Probability
Grade 12
Question:
<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 4:</strong> \(P(E)\) is maximum when \(x\) equal to</p>
<p>5</p>
<p>3</p>
<p>4</p>
<p>6</p>
Step-by-Step Solution
Key Concept: Find the probability distribution of the sum by computing P(sum = x) for all possible values of x, recognizing that we must account for the non-uniform probability distributions on each die (3/6, 2/6, 1/6 for die A and 1/6, 2/6, 3/6 for die B).
<p><strong>Step 1:</strong> Set up probability distributions for each die.</p><p>Die A: P(1)=3/6, P(2)=2/6, P(3)=1/6</p><p>Die B: P(1)=1/6, P(2)=2/6, P(3)=3/6</p><p><strong>Step 2:</strong> Calculate P(E) for each possible sum x from 2 to 6.</p><p>P(sum=2) = P(A=1)×P(B=1) = (3/6)×(1/6) = 3/36</p><p>P(sum=3) = P(A=1)×P(B=2) + P(A=2)×P(B=1) = (3/6)×(2/6) + (2/6)×(1/6) = 6/36 + 2/36 = 8/36</p><p>P(sum=4) = P(A=1)×P(B=3) + P(A=2)×P(B=2) + P(A=3)×P(B=1) = (3/6)×(3/6) + (2/6)×(2/6) + (1/6)×(1/6) = 9/36 + 4/36 + 1/36 = 14/36</p><p>P(sum=5) = P(A=2)×P(B=3) + P(A=3)×P(B=2) = (2/6)×(3/6) + (1/6)×(2/6) = 6/36 + 2/36 = 8/36</p><p>P(sum=6) = P(A=3)×P(B=3) = (1/6)×(3/6) = 3/36</p><p><strong>Step 3:</strong> Compare probabilities: 3/36 < 8/36 < 14/36 > 8/36 > 3/36</p><p>Maximum probability is 14/36 when x = 4.</p><p>∴ Answer: C (x = 4)</p>
Correct Answer: C