We need to find the conditional probability. The condition is that the sum of the numbers appearing on two dice casts is 10. We want to find the probability that the number 5 appeared at least once, given this condition.
Let A be the event that the sum of the numbers is 10. The possible outcomes $(x, y)$ where $x$ is the result of the first cast and $y$ is the result of the second cast are:
The total number of outcomes where the sum is 10 is $n(A) = 3$. This is our reduced sample space.
Let B be the event that the number 5 appears at least once.
We need to find the outcomes in event A (sum is 10) where the number 5 appears at least once. Looking at the outcomes in A:
The only outcome where the sum is 10 AND the number 5 appears at least once is (5, 5). So, the number of favorable outcomes is $n(A \cap B) = 1$.
The conditional probability $P(B|A)$ is calculated as:
$ P(B|A) = \frac{n(A \cap B)}{n(A)} $Substituting the values we found:
$ P(B|A) = \frac{1}{3} $Therefore, the probability that the number 5 has appeared at least once, given that the sum is 10, is $\frac{1}{3}$.
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