Given:
Pr[X]=0.6
Pr[Y]=0.4
Pr[A|X]=0.8
Pr[B|X]=0.2
Pr[A|Y]=0.3
Pr[B|Y]=0.4
Pr[C|Y]=0.3
Find the following missing probabilities:
(1) Pr[A]= .48+.12 (Correct)
(2) Pr[X|A]= .48/1.1 (Wrong)
(3) Pr[Y|B]= .16/.6 (Wrong)
(4) Pr[B|Y]= .4 (Correct)
So what I've currently done is made a tree diagram. From the tree diagram I found that, when X is chosen first, the Pr[A] is .8 and Pr[B] is .2. Meaning Pr[X intersect A] is .48 and Pr[X intersect B] is .12.
Next, when Y is chosen first, the Pr[A] is .3, Pr[B] is .4, and Pr[.3] is .3. This means Pr[Y intersect A] is .12, Pr[Y intersect B] is .16, and Pr[Y intersect C] is .12.
Now my problem is with problem 2 and 3. What I've done is for number I got .48 because it's Pr[X intersect A]. And with the denominator, I added the Pr[A] of both A giving me 1.1. The same idea with problem 3. Can someone tell me what's wrong?
\begin{align} \mathbb{P}[X|A] &= \frac{\mathbb{P}[A|X] \times \mathbb{P}[X]}{\mathbb{P}[A]} \\ &= \frac{\mathbb{P}[A|X] \times \mathbb{P}[X]}{\mathbb{P}[A|X] \times \mathbb{P}[X] + \mathbb{P}[A|Y] \times \mathbb{P}[Y]} \\ &= \frac{0.48}{0.48 + 0.12} = 0.8 \end{align}
\begin{align} \mathbb{P}[Y|B] &= \frac{\mathbb{P}[B|Y] \times \mathbb{P}[Y]}{\mathbb{P}[B]} \\ &= \frac{\mathbb{P}[B|Y] \times \mathbb{P}[Y]}{\mathbb{P}[B|Y] \times \mathbb{P}[Y] + \mathbb{P}[B|X] \times \mathbb{P}[X]} \\ &= \frac{0.16}{0.16 + 0.12} \approx 0.57 \end{align}