I have the following Bayesian network:
Given all the values of P(X1), P(X2), P(X3|X1), P(X4|X1,X2), how would you find P(X4=1|X3=1)?
So far, I have tried the following manipulation with product and chain rule but I believe it is wrong and I am unsure how else to go about this:
$$P(X4|X3)=\frac{P(X4,X3)}{P(X3)}=\frac{=\prod_{} P(X3|X1)P(X4|X1,X2)P(X1)P(X2)}{P(X3)}$$
Assuming the possible values of each random variable is $0$ or $1$ and the implications of the Bayesian network, you may get $$P(X_4=1 \mid X_3=1) = \dfrac{P(X_3=1,X_4=1)}{P(X_3=1)}$$
where
$$P(X_3=1) = P(X_3=1\mid X_1=1)P(X_1=1) + P(X_3=1\mid X_1=0)P(X_1=0)$$
and $$P(X_3=1,X_4=1) \\= P(X_3=1,X_4=1 \mid X_1=1,X_2=1)P(X_1=1,X_2=1) \\+P(X_3=1,X_4=1 \mid X_1=1,X_2=0)P(X_1=1,X_2=0) \\+ P(X_3=1,X_4=1 \mid X_1=0,X_2=1)P(X_1=0,X_2=1) \\+P(X_3=1,X_4=1 \mid X_1=0,X_2=0)P(X_1=0,X_2=0) \\= P(X_3=1\mid X_1=1)P(X_4=1 \mid X_1=1,X_2=1)P(X_1=1)P(X_2=1) \\+ P(X_3=1\mid X_1=1)P(X_4=1 \mid X_1=1,X_2=0)P(X_1=1)P(X_2=0) \\+ P(X_3=1\mid X_1=0)P(X_4=1 \mid X_1=0,X_2=1)P(X_1=0)P(X_2=1) \\+ P(X_3=1\mid X_1=0)P(X_4=1 \mid X_1=0,X_2=0)P(X_1=0)P(X_2=0) $$