I'm struggling trying to calculate the probabilities associated with a random variable $Z$ whose value depends on the realizations of two other random variables $X,Y$. I have :
$$
Pr(X=A)=0.25, Pr(Y=D|X=A)=0.60,Pr(Y=E|X=A)=0.30, Pr(Y=F|X=A)=0.10\\
Pr(X=B)=0.50, Pr(Y=D|X=B)=0.40,Pr(Y=E|X=B)=0.30, Pr(Y=F|X=B)=0.30\\
Pr(X=C)=0.25, Pr(Y=D|X=C)=0.10,Pr(Y=E|X=C)=0.20, Pr(Y=F|X=C)=0.70
$$
The probability distribution of $Y$ depends on the value taken by $X$. The random variable $Z$ takes the following values depending on the values taken by $\{X,Y\}$:
$$
\begin{aligned}
\{X,Y\} &= \{A,F\} &&\rightarrow Z=20\\
&= \{B,F\} &&\rightarrow Z=18\\
&= \{A,E\} &&\rightarrow Z=17\\
&= \{C,F\} &&\rightarrow Z=16\\
&= \{A,D\} \text{ or } \{B,E\} &&\rightarrow Z=15\\
&= \{B,D\} \text{ or } \{C,E\} &&\rightarrow Z=13\\
&= \{C,D\} &&\rightarrow Z=11\\
\end{aligned}
$$
I am interested in the $Pr(Z=i) : i \in \{11,13,15,16,17,18,20\}$.
1st try : for $Z=15$
$$
\begin{aligned}
Pr(Z=15) &= Pr(A \cap D) \cup Pr(B \cap E) = (0.60)(0.25) + (0.30)(0.50) = 0.30
\end{aligned}
$$
2nd try :
$$
\begin{aligned}
Pr(Z=15) &= Pr(D|A)\cdot Pr(A) + Pr((E \cap B) | A^{c}) \cdot Pr( A^{c} )\\
\end{aligned}
$$
However, I'm not sure of how to calculate the conditional probability $Pr((E \cap B) | A^{c})$ on this second approach.
It looks like this "simple" question is taking me too long to solve, so what I'm I doing so wrong?
The correct answer is :
$$
\begin{aligned}
\{X,Y\} &= \{A,F\} &&\rightarrow P(Z=20)=0.075\\
&= \{B,F\} &&\rightarrow P(Z=18)=0.15\\
&= \{A,E\} &&\rightarrow P(Z=17)=0.075\\
&= \{C,F\} &&\rightarrow P(Z=16)=0.075\\
&= \{A,D\} \text{ or } \{B,E\} &&\rightarrow P(Z=15)=0.25\\
&= \{B,D\} \text{ or } \{C,E\} &&\rightarrow P(Z=13)=0.275\\
&= \{C,D\} &&\rightarrow P(Z=11)=0.10\\
\end{aligned}
$$
Best Answer
$P(Z=20) = P(X=A, Y=F) = P(X=A) P(Y=F|X=A) = 0.25 \, \cdot \, 0.1$, so there must be something wrong with the values you have given - either in the definition, or in the "correct" ones.