# What is the probability of two dice getting a sum of 7 without a two?

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 $P(A|B)$.

• Why does it have to be conditional? $7=6+1=1+6=3+4=4+3$ How many are there? – Eleven-Eleven May 3 '15 at 22:26
• What do think A and B should be - what events? – ip6 May 3 '15 at 22:27
• 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. – Eleven-Eleven May 3 '15 at 22:32
• @ampage, can you be a little more specific as to what you are looking for? – Eleven-Eleven May 3 '15 at 22:48
• @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. – Ampage Green May 3 '15 at 23:46

$B= \{1,3,4,5,6\}^2$.
$A= \{ (i,j) | i+j = 7, i, j \in \{1,2,3,4,5,6\} \}$.
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 $36$, $P(E)=4/36=1/9$.
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, $$P(x_1 \ne 2 \cap x_2\ne 2 \cap x_1 + x_2 = 7) = P(x_1\ne 2 \cap x_2 \ne 2 | x_1 + x_2 = 7){P(x_1 + x_2 = 7)}.$$