A fair die is rolled three times independently. Given that 6 appeared at least once, the conditional probability that 6 appeared exactly twice equals ..............
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Correct Answer: 0.16 - 0.17
Solution and Explanation
Step 1: Define the total number of possible outcomes.
When a fair die is rolled three times, the total number of outcomes is \( 6^3 = 216 \).
Step 2: Calculate the conditional probability.
We are given that at least one 6 appeared. The total number of outcomes where at least one 6 appears can be calculated by subtracting the outcomes where no 6 appears from the total outcomes: \[ \text{Outcomes with at least one 6} = 6^3 - 5^3 = 216 - 125 = 91. \]
Step 3: Calculate the number of outcomes with exactly two 6s.
To have exactly two 6s, we must choose 2 positions out of 3 for the 6s, and the remaining position must be any of the 5 other numbers: \[ \text{Outcomes with exactly two 6s} = \binom{3}{2} \times 5 = 3 \times 5 = 15. \]
Step 4: Calculate the conditional probability.
The conditional probability is the ratio of outcomes with exactly two 6s to the total outcomes with at least one 6: \[ P(\text{exactly two 6s} \mid \text{at least one 6}) = \frac{15}{91} = 0.164 \]
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