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I am currently working on conditional probability and I am somewhat confused about how exactly to complete this problem. I know that to find conditional probability that you utilize:

$$P(A|B) = \frac{P(A\cap B)}{P(B)}$$

I also know that there is a $6/36$ chance to roll a sum of 7, and that if you roll a sum of 7 that there is a $4/6$ chance to get a sum without using the number 2. I do not know what else is necessary however in order to finish this problem and to find $P(A|B)$.

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    $\begingroup$ Why does it have to be conditional? $7=6+1=1+6=3+4=4+3$ How many are there? $\endgroup$ – Eleven-Eleven May 3 '15 at 22:26
  • $\begingroup$ What do think A and B should be - what events? $\endgroup$ – ip6 May 3 '15 at 22:27
  • $\begingroup$ Wouldn't it just be $1/9$? I'm reading it as you can roll a seven using two fair die including 2's, you just are not looking for specifically 5+2 or 2+5 as part of your event. $\endgroup$ – Eleven-Eleven May 3 '15 at 22:32
  • $\begingroup$ @ampage, can you be a little more specific as to what you are looking for? $\endgroup$ – Eleven-Eleven May 3 '15 at 22:48
  • $\begingroup$ @Eleven-Eleven The question asks "A pair of dice are rolled. What is the probability that neither die shows a 2 given that they sum to 7? What is the probability that they sum to 7 given that neither die shows a 2?" It seems to me that the answer would be the same, however, I'm not sure why it would be asked inversely if that were the case. $\endgroup$ – Ampage Green May 3 '15 at 23:46
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$B= \{1,3,4,5,6\}^2$.

$A= \{ (i,j) | i+j = 7, i, j \in \{1,2,3,4,5,6\} \}$.

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The Event $E$of rolling two dice that sum to seven not including a sum with summand 2 would be $$\{(1,6),(3,4),(4,3),(6,1)\}$$

Thus $|E|=4$. Since the sample space has cardinality $36$, $P(E)=4/36=1/9$.

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I also know that there is a 6/36 chance to roll a sum of 7, and that if you roll a sum of 7 that there is a 4/6 chance to get a sum without using the number 2.

So, putting these statements into probability notation, you have, $$ \begin{split} P(x_1 + x_2 = 7) &= \frac{6}{36}\\ P(x_1\ne 2 \cap x_2 \ne 2 | x_1 + x_2 = 7) &= \:\frac{4}{6} \end{split} $$ So, what you want is $P(x_1 \ne 2 \cap x_2\ne 2 \cap x_1 + x_2 = 7)$ (i.e. the joint probability), so using the conditional probability definition: $$ P(x_1\ne 2 \cap x_2 \ne 2 | x_1 + x_2 = 7) = \frac{P(x_1 \ne 2 \cap x_2\ne 2 \cap x_1 + x_2 = 7)}{P(x_1 + x_2 = 7)} \qquad $$ and thus, $$ P(x_1 \ne 2 \cap x_2\ne 2 \cap x_1 + x_2 = 7) = P(x_1\ne 2 \cap x_2 \ne 2 | x_1 + x_2 = 7){P(x_1 + x_2 = 7)}. $$

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