A die was thrown twice and it was found that the sum of the numbers that appeared was 6. Find the conditional probability that the number 4 appeared at least once.
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Solution and Explanation
Step 1: The possible outcomes when a die is thrown twice and the sum is 6 are: \[ (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) \] So, there are 5 outcomes in total. Step 2: The favorable outcomes for the number 4 to appear at least once are: \[ (2, 4), (4, 2) \] So, there are 2 favorable outcomes.
Step 3: The conditional probability is given by: \[ P({4 appears at least once}) = \frac{{Number of favorable outcomes}}{{Total number of outcomes}} = \frac{2}{5} \] Thus, the conditional probability is \( \frac{2}{5} \).
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