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

Question

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 $P(A|B)$.

Answer 1

$B= \{1,3,4,5,6\}^2$.

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

Answer 2

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 $36$, $P(E)=4/36=1/9$.

Answer 3

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, $$ P(x_1 \ne 2 \cap x_2\ne 2 \cap x_1 + x_2 = 7) = P(x_1\ne 2 \cap x_2 \ne 2 | x_1 + x_2 = 7){P(x_1 + x_2 = 7)}. $$