Question

In a single throw of pair of dice, the probability of getting a multiple of 2 on one and a multiple of 3 on the other will be :

• A. \(\frac{5}{36}\)
• B. \(\frac{1}{18}\)
• C. \(\frac{11}{36}\)
• D. \(\frac{1}{6}\)

Hint:

Total outcomes with two dice = 36. Multiples of 2 (M2): {2, 4, 6}. Multiples of 3 (M3): {3, 6}. We need (M2 on 1st die AND M3 on 2nd die) OR (M3 on 1st die AND M2 on 2nd die). Pairs for (M2, M3): (2,3), (2,6), (4,3), (4,6), (6,3), (6,6) - 6 pairs. Pairs for (M3, M2): (3,2), (3,4), (3,6), (6,2), (6,4), (6,6) - 6 pairs. The pair (6,6) is common. So, unique favorable pairs = \(6+6-1 = 11\). Probability = \(11/36\).

Answer 1

The Correct Option is C

Concept: Probability = (Favorable Outcomes) / (Total Outcomes). Total outcomes for two dice = \(6 \times 6 = 36\). Step 1: Define events Let D1 be the outcome of the first die, D2 for the second. M2 = Multiple of 2 = \{2, 4, 6\} (3 outcomes) M3 = Multiple of 3 = \{3, 6\} (2 outcomes) We want (D1 \(\in\) M2 AND D2 \(\in\) M3) OR (D1 \(\in\) M3 AND D2 \(\in\) M2). Step 2: List favorable pairs Case A: (D1 is M2, D2 is M3) Pairs: (2,3), (2,6), (4,3), (4,6), (6,3), (6,6). (Count = \(3 \times 2 = 6\) pairs) Case B: (D1 is M3, D2 is M2) Pairs: (3,2), (3,4), (3,6), (6,2), (6,4), (6,6). (Count = \(2 \times 3 = 6\) pairs) Step 3: Combine cases and remove overlap The outcome (6,6) is in both Case A and Case B (6 is a multiple of 2 and 6 is a multiple of 3). Total unique favorable outcomes = (Outcomes in A) + (Outcomes in B) - (Overlapping outcomes) Total = \(6 + 6 - 1\) (since (6,6) is the only overlap). Total favorable outcomes = 11. Step 4: Calculate probability Probability = \(\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{11}{36}\).