Question

Two dice are rolled. If \( A \) denotes the event that the same number shows on each die and \( B \) denotes the event that the sum of the numbers on both dice is greater than 7, then \( P(A \mid B) \) and \( P(B \mid A) \) respectively are:

• A. \(\dfrac{2}{5}, \dfrac{1}{4}\)
• B. \(\dfrac{1}{5}, \dfrac{1}{2}\)
• C. \(\dfrac{1}{5}, \dfrac{1}{4}\)
• D. \(\dfrac{1}{2}, \dfrac{3}{5}\)

Hint:

Conditional probability \( P(A \mid B) \) is calculated as \( \frac{P(A \cap B)}{P(B)} \). Carefully identify the outcomes for each event and their intersection.

Answer 1

The Correct Option is B

Total outcomes when two dice are rolled: \[ n(S) = 6 \times 6 = 36 \] Let us define the events: - \( A \): same number on both dice → outcomes: \[ \{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\}, \quad \Rightarrow P(A) = \dfrac{6}{36} = \dfrac{1}{6} \] - \( B \): sum greater than 7 → valid sums: 8, 9, 10, 11, 12 These correspond to the following outcomes: - 8: (2,6), (3,5), (4,4), (5,3), (6,2) - 9: (3,6), (4,5), (5,4), (6,3) - 10: (4,6), (5,5), (6,4) - 11: (5,6), (6,5) - 12: (6,6) So total outcomes in \( B \): 15 \[ P(B) = \dfrac{15}{36} = \dfrac{5}{12} \] Now, find: - \( A \cap B \): both same number and sum>7 This includes: (4,4), (5,5), (6,6) → 3 outcomes \[ P(A \cap B) = \dfrac{3}{36} = \dfrac{1}{12} \] Then: - \( P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{1/12}{5/12} = \dfrac{1}{5} \) - \( P(B \mid A) = \dfrac{P(A \cap B)}{P(A)} = \dfrac{1/12}{1/6} = \dfrac{1}{2} \) % Final Answer \[ \boxed{P(A \mid B) = \dfrac{1}{5}, \quad P(B \mid A) = \dfrac{1}{2}} \]