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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.

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1 Answer 1

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'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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  • $\begingroup$ Yes I knew that the answer is identical to P(A and B). But how do you solve that with the formula p(A|B) = P(A and B)/P(B)? $\endgroup$
    – Gtexx
    Commented Nov 5, 2020 at 10:17
  • $\begingroup$ @Gtexx : $\frac{P(A \cap B)}{P(B)}=P(A)$ but if you want to answer the question you put you have to do $P(B|A)=P(B)=0.5$. In the meantime I would like to encourage you to click the check mark if this answer was helpful for you. It'll encourage others to answer your questions in the future $\endgroup$
    – tommik
    Commented Nov 5, 2020 at 10:21
  • $\begingroup$ I have ticked your answer. I get it that P(A and B) = 15/72. But how do you prove that P(A and B) = P(A|B) = P(A)? I know I am asking too many questions, but please I am just trying to master the conditional probability. $\endgroup$
    – Gtexx
    Commented Nov 5, 2020 at 10:38
  • $\begingroup$ @Gtexx: $P(A \cap B)=P(A)P(B)$ per independence $\endgroup$
    – tommik
    Commented Nov 5, 2020 at 10:41
  • $\begingroup$ P(A and B) / P(B) = (15/72) / 1/2 = 0.416. This does not equal 0.208 $\endgroup$
    – Gtexx
    Commented Nov 5, 2020 at 10:45

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