Question

Two dice are thrown simultaneously. The probability of getting a multiple of \(2\) on one die and a multiple of \(3\) on the other die is:

• A. \(\frac{5}{12}\)
• B. \(\frac{11}{36}\)
• C. \(\frac{13}{36}\)
• D. \(\frac{5}{36}\)

Hint:

When a probability question involves “one die” and “the other die”, count both possible arrangements and then subtract the overlap using the inclusion-exclusion principle.

Answer 1

The Correct Option is B

Step 1: Calculate the total number of outcomes for two dice.
Total outcomes = 6 × 6 = 36.

Step 2: Define the events clearly.
Let the requirement be that one die shows a multiple of 2 and the other shows a multiple of 3. Multiples of 2 are {2, 4, 6} (3 choices). Multiples of 3 are {3, 6} (2 choices).

Step 3: Count outcomes for both ordered arrangements.
First die multiple of 2, second multiple of 3: 3 × 2 = 6 outcomes. First die multiple of 3, second multiple of 2: 2 × 3 = 6 outcomes. Preliminary total = 12.

Step 4: Remove the doubly-counted intersection.
The outcome (6, 6) satisfies both arrangements simultaneously and was counted twice. Subtract this one overlap: 12 - 1 = 11 favorable outcomes.

Step 5: Compute the probability.
P = 11/36.

Step 6: Final conclusion.
The required probability is 11/36.