Let X and Y be the Random variables that describe pressure on car wheels (X refers to the front wheel, Y refers to the back wheel). They follow a probability distribution according to the table:
| X/Y | 10 | 12 | 14 |
|---|---|---|---|
| 13 | .1229 | .0875 | .1229 |
| 14 | .1229 | .1229 | .0875 |
| 15 | .0875 | .1229 | .1229 |
What is the expected value of $X - Y$ conditional at $X = 13$?
My guess is to do $$P(X = 13) = P(X = 10, Y = 10) + P(X = 10, Y = 12) + P(X = 10, Y = 14)$$ and then I have $$P(Y = y \,|\, X = 13) = \frac{P(X = 13, Y = y)}{P(X = 13)}~.$$
But from now how do I get to the expected value of $Y | X = 13$?
As was pointed out in a comment you can compute $\mathbb{E}(Y|X=13)$ by iterating over all the possible values for Y and multiplying by the probability of each occurring given $X=13$
$$\mathbb{E}(X|Y=13) = 10\mathbb{P}(Y=10∣X=13)+12\mathbb{P}(Y=12∣X=13)+14\mathbb{P}(Y=14∣X=13) $$
To get $\mathbb{P}(Y=10∣X=13)$, for example, we can use \begin{align*} \mathbb{P}(Y=10∣X=13)&=\frac{\mathbb{P}(Y=10,X=13)}{\mathbb{P}(X=13)}\\ &=\frac{0.1229}{0.1229 + 0.0875 + 0.1229}\\ &=0.5372 \end{align*}
Doing similar for $\mathbb{P}(Y=12∣X=13)$ and $\mathbb{P}(Y=14∣X=13)$ and plugging into the first equation will give us \begin{align*} \mathbb{E}(X|Y=13) &= 10\mathbb{P}(Y=10∣X=13)+12\mathbb{P}(Y=12∣X=13)+14\mathbb{P}(Y=14∣X=13)\\ &=10*0.5372 + 12*0.3824 + 14*0.5372\\ &=17.4815 \end{align*}