This problem involves conditional probability. We need to find the probability of an event (getting at least one 5) given that another event has already occurred (the sum of the numbers is 10).
First, let's list all possible pairs of outcomes when rolling a dice twice that sum up to 10. Let the outcome be represented as (roll1, roll2).
There are 3 possible outcomes where the sum is 10. This is our reduced sample space.
Next, we identify which of these outcomes include the number 5 appearing at least once.
Only 1 outcome, (5, 5), satisfies the condition of having at least one 5.
The probability is the ratio of the number of favorable outcomes to the total number of possible outcomes in the reduced sample space.
Let A be the event that the number 5 appears at least once.
Let B be the event that the sum of the numbers is 10.
We are looking for P(A | B).
Number of outcomes where sum is 10 = 3.
Number of outcomes where sum is 10 AND at least one 5 appears = 1 (which is the outcome (5, 5)).
The conditional probability is calculated as: $ P(\text{At least one 5} | \text{Sum is 10}) = \frac{\text{Number of outcomes with sum 10 and at least one 5}}{\text{Number of outcomes with sum 10}} $
$ P(A | B) = \frac{1}{3} $Three dice are thrown. What is the probability of getting a sum which is a perfect square?
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