The first die is already 6, so to satisfy the condition that sum is greater than 8, the second die must be {3,4,5,6}. We have 4 events that satisfy the condition, and 6 possible events (getting any number from 1 to 6). So the probability is
"\\displaystyle P(A) = \\frac{4}{6} = \\frac{2}{3}"
Answer: 2/3
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