Bayes theorem. Conditional ball draws You have a white ball in your hand and  there is a ball in a black box that has a probability of 50% of being   white. You put the white ball you are   holding in the box and shake up the box.
a. What is the chance of pulling out a white ball   from    the box?
b. You  take out the white ball. What is the probability of the remaining ball being white?
PROPOSED SOLUTION:
Let white ball in hand is Ball-1 and other ball in box is Ball-2
A - Ball-1 is pulled out
B - Ball-2 is pulled out
W - First draw is white
X - Second draw is white.
(a) Now,
P(W) = P(W|A)P(A) + P(W|B)P(B) = 1/2*1 + 1/2*1/2 = 3/4
(b) P(X|W) = P(X)P(W|X)/P(W)
Not sure how to proceed from here. Any ideas?
 A: (a) $\mathsf P(W) = \mathsf P(W\mid A)\,\mathsf P(A)+\mathsf P(W\mid B)\,\mathsf P(B)$ is okay. $\color{blue}\checkmark$
( Note use a vertical slash rather than diagonal to avoid confusion with division. )
(b) Second verse should be the same as the first; just a slight adjustment for the conditional.
$\qquad\mathsf P(X\mid W) = \mathsf P(X\mid W, A)\,\mathsf P(A\mid W)+\mathsf P(X\mid W, B)\,\mathsf P(B \mid W)$
And of course, $\mathsf P(A\mid W) = \mathsf P(W\mid A)\,\mathsf P(A)\Big/\mathsf P(W)$ etc., so:
$\qquad\mathsf P(X\mid W) = \tfrac 2 3 \mathsf P(X\mid W, A)+\frac 1 3 \mathsf P(X\mid W, B)$
So what is the probability that the second ball is white given that ball-1 was pulled out first and was white?   $\mathsf P(X\mid W, A)=?$
So what is the probability that the second ball is white given that ball-2 was pulled out first and was white?   $\mathsf P(X\mid W, B)=?$


$\mathsf P(X\mid W,A)=1/2$ and $\mathsf P(X\mid W,\color{red}{B})=1$ – Rahul

Exactly (assuming that second $A$ was a cut-and-paste-error).

Thanks Graham. Could you please explain: $\mathsf P(X\mid W)=\mathsf P(X\mid W,A)\,\mathsf P(A\mid W)+\mathsf P(X\mid W,B)\,\mathsf P(B\mid W)$ and why it is not $\mathsf P(X\mid W)=\mathsf P(X\mid W,A)\,\mathsf P(A)+\mathsf P(X\mid W,B)\,\mathsf P(B)$ – Rahul

Certainly.   We use Bayes' Rule.
$$\begin{align}
\mathsf P(X\mid W) & = \dfrac{\mathsf P(X, W)}{\mathsf P(W)}
\\[1ex] & = \dfrac{\mathsf P(X, W, A)+\mathsf P(X, W, B)}{\mathsf P(W)}
\\[1ex] & = \dfrac{\mathsf P(X\mid W, A)\,\mathsf P(W,A)+\mathsf P(X\mid W, B)\,\mathsf P(W, B)}{\mathsf P(W)}
\\[1ex] & = \mathsf P(X\mid W, A)\,\mathsf P(A\mid W)+\mathsf P(X\mid W, B)\,\mathsf P(B\mid W)
\\[4ex] & = \mathsf P(X\mid W, A)\,\mathsf P(A)\,\frac{\mathsf P(W\mid A)}{\mathsf P(W)}+\mathsf P(X\mid W, B)\,\mathsf P(B)\,\frac{\mathsf P(W \mid B)}{\mathsf P(W)}
\end{align}$$
