Given random variables X, Y, and Z, each of which with domain {-1, 1}, compute $P(A=1|B=1 \cap C=1)$ given that the following are true:
- $P(X=1)=0.55$
- $P(Y=1|X=1)=0.66$
- $P(Y=1|X=-1)=0.2$
- $P(Z=1|X=1)=0.94$
- $P(Z=1|X=-1)=0.12$
- Z and Y are conditionally independent given X.
How do you compute this? I assume I'm supposed to use the chain rule, but:
$$P(X=1|Y=1 \cap Z=1)=\frac{P(X=1\cap Y=1\cap Z=1)}{P(Y=1 \cap Z=1)}$$
doesn't really help me much since I don't know the probability of any of these intersections.
After the original post was edited, there appears additional information: $$ \mathbb P(Y=1,Z=1\mid X=1)=\mathbb P(Y=1\mid X=1)\cdot \mathbb P(Z=1\mid X=1)=0.94\cdot 0.66 $$ and $$ \mathbb P(Y=1,Z=1\mid X=-1)=\mathbb P(Y=1\mid X=-1)\cdot \mathbb P(Z=1\mid X=-1)=0.12\cdot 0.2 $$
From the first equality you can find $\mathbb (X=1,Y=1,Z=1)$.
To find $\mathbb P(Y=1, Z=1)$ find first $\mathbb (X=-1,Y=1,Z=1)$ from the second equality above and add $$ \mathbb P(Y=1, Z=1) = \mathbb P(X=-1,Y=1, Z=1)+\mathbb P(X=1,Y=1, Z=1). $$