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A fun alternative way to solve:
Event A: sum of 5
Lets assume you firstly throw the one dice, and then the other. In order to have sum 5, the first dice must NOT be 5 or 6. Probability: 4/6. If the first dice is 1, 2, 3 or 4 then there is only ONE number that can satisfy the condition (4, 3, 2, 1 respectively) "sum of 5" (the numbers are 4, 3, 2, 1 respectively). this makes is a total of: (4/6)*(1/6)=4/36 probability.
Event B: rolling doubles (ex. 3, 3)
Again, you throw the first dice. In this case, we don't care about the results. In the 2nd throw, there is only ONE number that satisfies the condition (1/6) (for example, if the 1st dice is 5, then there is 1/6 probability for the 2nd dice to be 5).
Find P (A and B) If I understand correctly, P is the probability of either A or B to be true.
If you pay attention, you will see that events A and B are INDEPENDENT. There is no way to have sum 5 and to roll doubles (rolling doubles will get you an even sum). Because of this, you can easily calculate the total probability (of A and B) by adding the numbers: P=4/36+1/6=6/36+6/36=10/36.
EDIT: Now that I think about it, P should be the probability of A and B to occur the same time. If that is the case:
A and B cannot occur the same time. Rolling doubles means you will have an even sum. So, your sum can't be 5. Because of this: P=0.
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