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Question
A pair of dice is thrown. Find the probability of getting the sum 8 or more, if 4 appears on the first die.
Solution
Consider the given events.
A = 4 appears on first die
B = The sum of the numbers on two dice is 8 or more.
Clearly,
A = {(4, 1), (4, 2), (4, 3), (4, 4) (4, 5), (4, 6)}
n(A) = 6
B = {(2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6),(5, 3), (5, 4), (5, 5) (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(B) = 15
\[\text{ Now } , \]
\[A \cap B = \left\{ \left( 4, 4 \right), \left( 4, 5 \right), \left( 4, 6 \right) \right\}\]
\[ \therefore \text{ Required probability } = P\left( B/A \right) = \frac{n\left( A \cap B \right)}{n\left( A \right)} = \frac{3}{6} = \frac{1}{2}\]
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