Two dice are tossed once. The probability of getting an even number at the first die or a total of \(8\) is

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Two dice are tossed once. The probability of getting an even number at the first die or a total of \(8\) is

A. \(\dfrac1{36}\)

B. \(\dfrac3{36}\)

C. \(\dfrac{11}{36}\)

D. \(\dfrac{20}{36}\)

E. \(\dfrac{23}{36}\)

Answer: D

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This does not look like an official question to me, at least if it has been transcribed correctly.

The probability the first die is even is 1/2, so since we want the probability either that happens or something else happens, the answer must be at least 1/2, and only D or E could possibly be right.

When we roll two dice in order, 6*6 = 36 things can happen. On half of those, or on 18 of those rolls, the first die is even. If we list the ways we might get a sum of 8, listing the two dice in order, we have:

2, 6
3, 5
4, 4
5, 3
6, 2

We've already counted three of these possibilities because in three of them, the first die is even. It's only the rolls (3, 5) and (5, 3) we haven't yet counted. So adding those 2 possibilities to the 18 where the first die is even, we have 20 ways to get the result the question asks about, out of 36 possibilities in total, and D is the answer.
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