Question

A pair of dice is thrown. Find the probability of getting 7 as the sum, if it is known that the second die always exhibits an odd number.

Answer 1

Consider the given events.
A = Number appearing on second die is odd
B = The sum of the numbers on two dice is 7.

Clearly,
A = {(1, 1), (1, 3),(1, 5), (2, 1), (2, 3), (2, 5), (3, 1), (3, 3), (3, 5), (4, 1), (4, 3), (4, 5), (5, 1), (5, 3), (5, 5),(6, 1), (6, 3), (6, 5)}
B = {(2, 5), (5, 2), (3, 4), (4, 3), (1, 6), (6, 1)}

\[\text{ Now } , \]

\[A \cap B = \left\{ \left( 2, 5 \right), \left( 4, 3 \right), \left( 6, 1 \right) \right\}\]

\[ \therefore \text{ Required probability}  = P\left( B/A \right) = \frac{n\left( A \cap B \right)}{n\left( A \right)} = \frac{3}{18} = \frac{1}{6}\]