Find $P(XY+Z-XZ=2)$ when $Z$~$N(0,1)$, $Y$~$B(3, \frac{1}{2})$, $X$~$B(1, \frac{1}{3})$

Question

Let $X,Y,Z$ be independent variables such that: \begin{gather*} Z \sim N(0,1) \,\text{(normal distribution)},\\ Y \sim B(3, \frac{1}{2}) \,\text{(binomial distribution)},\\ X \sim B(1, \frac{1}{3}) \,\text{(binomial distribution)}. \end{gather*}

We define $W=XY+(1-X)Z$. I need to calculate $P(X=1|W=2)$ , so\begin{align*} P(X=1|W=2)&=P(X=1|XY+Z-XZ=2)\\ &=\frac{P(XY+Z-XZ=2 \cap X=1)}{P(XY+Z-XZ=2)}\\ &= \frac{P(Y=2)}{P(XY+Z-XZ=2)}. \end{align*} And I got stuck in calculating $P(XY+Z-XZ=2)$. Any help would be appreciated.

Answer 1

There is a more simple route.

Note that: $X=0\wedge W=2\implies Z=2$ so that $P(X=0\wedge W=2)\leq P(Z=2)=0$.

Then: $$P(W=2)=P(X=0\wedge W=2)+P(X=1\wedge W=2)=P(X=1\wedge W=2)$$

so that $$P(X=1\mid W=2)=P(X=1\wedge W=2)/P(W=2)=1$$

Answer 2

\begin{align*} P(XY + Z - XZ = 2) &= P(XY + Z - XZ = 2 \mid X = 0) + P(XY + Z - XZ = 2 \mid X = 1)\\ &= P(Z = 2 \mid X = 0) + P(Y = 2 \mid X = 1)\\ &= P(Z = 2) P(X = 0) + P(Y = 2)P(X = 1)\\ &= 0 \cdot \frac{2}{3} + \frac{3}{8} \cdot \frac{1}{3} = \frac{1}{8}. \end{align*}