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