# conditional probability of dependent random variables

Suppose I have 3 random variables:

$$X \sim \mbox{Bernoulli}(1/2)$$ $$Z \sim \mbox{Normal}(0,1)$$ $$Y = X+Z$$

How do I compute the conditional probability:

$$P(X=1 | Y=y)$$

# Attempt1:

Probability[ X == 1 \[Conditioned] X + Z == y,             {             X \[Distributed] BernoulliDistribution[1/2]            ,Z \[Distributed] NormalDistribution[]            }          ] 

# Attempt2:

D[Probability[ X == 1 \[Conditioned] X + Z >= y,             {             X \[Distributed] BernoulliDistribution[1/2]            ,Z \[Distributed] NormalDistribution[]            }          ],y] 

# Attempt3:

Likelihood[       TransformedDistribution[X + Z,                     {                     X \[Distributed]BernoulliDistribution[1/2],                     Z \[Distributed] NormalDistribution[]}]            , {y}] 

# Pencil and Paper attempt:

$$P(X=1 | Y=y) = \frac{P(X=1 , Y=y)}{P(Y=y)}$$ $$= \frac{P(X=1 , X+Z=y)}{P(Y=y)}$$ $$= \frac{P(X=1)P(Z=y-1)}{P(Y=y)}$$ $$= \frac{P(X=1)P(Z=y-1)}{P(X=1)P(Z=y-1)+P(X=0)P(Z=y-0)}$$

$$P(Z=y)=\frac{e^{-\frac{y^2}{2}}}{\sqrt{2 \pi }}$$ $$P(Z=y-0)=\frac{e^{-\frac{y^2}{2}}}{\sqrt{2 \pi }}$$ $$P(Z=y-1)=\frac{e^{-\frac{1}{2} (y-1)^2}}{\sqrt{2 \pi }}$$ $$P(X=1)=\frac{1}{2}$$ $$P(X=0)=\frac{1}{2}$$

$$P(X=1 | Y=y) = \frac{e^{-\frac{1}{2} (y-1)^2}}{2 \sqrt{2 \pi } \left(\frac{e^{-\frac{y^2}{2}}}{2 \sqrt{2 \pi }}+\frac{e^{-\frac{1}{2} (y-1)^2}}{2 \sqrt{2 \pi }}\right)}$$

$$P(X=1|Y=y) = \frac{e^y}{e^y+\sqrt{e}}$$