The probability that the second roll shows an even sum, given that the first roll shows a sum greater than 7

probability

As the title states: 'If two fair dice are rolled twice, what is the probability that the second roll shows an even sum, given that the first roll shows a sum greater than 7?'

My textbook's answer is 0.208
∧
My answer is 0.6, and my working out:

P(B|A) = P(A ∧ B)/P(A)

P(A ∧ B) = P(A)*P(B)

P(A) = 15/36 {(2,6), (3,5), (3,6), (4,4), (4,5), (4,6),(5,3), (5,4), (5,5), (5,6), (6,2), (6,3), (6,4), (6,5), (6,6)}

P(B) = 1/2

Therefore, 15/36 * 1/2 = 15/72

p(B|A) = P(A ∧ B)/P(A) = (15/72) / (15/36) => 15/72 * 36/15 => 1/2

I am not sure how the official answer is 0.208.

Please help me with this one.

Best Answer

'If two fair dice are rolled twice, what is the probability that the second roll shows an even sum, given that the first roll shows a sum greater than 7?'

As per independence, the probability that the second roll (second roll of the two dice) shows an even sum is always 0.5. This probability is not affected by the result of the previous two dice's roll.

The answer 0.208 of your textbook is just $\frac{15}{72}$, that can be, for example, the answer to the following similar question:

'If two fair dice are rolled twice, what is the probability that the second roll shows an even sum AND that the first roll showed a sum greater than 7?'

So probably the solution is referred to another exercise... but there is a certain confusion also in your reasoning.

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