Conditional Probability - Drawing balls from an urn

Question

I have that question from a past exam (without answer):

There are two urns, say I and II. Urn I contains 1 white ball and 1 black ball. Urn II containts two white balls and 3 black balls, and suppose that the balls are indistinguishable except for the colour. A ball is drawn randomly from urn I and put into urn II. Then a ball is drawn from urn II. Determine: (...) b) The probability of the 1st drawn ball being white, given that the 2nd ball was a white ball.

My answer:

$P(W_{1}|W_{2}) = \dfrac{P(W_{2}|W_{1})P(W_{1})}{P(W_{2})} = \dfrac{\Big(\dfrac{1}{2}.\dfrac{1}{2}\Big).\Big(\dfrac{1}{2}\Big)}{P(W_{2})} = \dfrac{\dfrac{1}{8}}{P(W_{2})}$

And here is my doubt: Should I consider $P(W_2) = 1$, as the condition is "white ball in the second draw" and get $P(W_{1}|W_{2}) = \dfrac{1}{8}$, or should I consider that $\text{urn }II'$ = $\begin{cases} 3W, 3B,\text{ if $W_{1}$}\\ 2W, 4B, \text{ if $B_{1}$} \end{cases}$, and then the probability of $W_{2}$ is $\dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6}$?

Answer 1

Firstly, $P(W_2|W_1)\neq \frac{1}{2}.\frac{1}{2}$ as in your numerator, because given the first ball is white, Urn 2 contains 3 white and 3 black balls, which yield $1/2$ for drawing a white ball. You shouldn't multiply with $1/2$ again. It is already multiplied in $P(W_1)$. So, your numerator is $1/4$.

And, irrespective of what is given in the question, $P(W_2)\neq 1$. $P(W_2)$ considers all the cases leading to a white ball in the second draw; which is $P(W_2)=P(W_2\cap W_1)+P(W_2\cap W_1')=\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}=\frac{5}{12}$.

Finally, $P(W_1|W_2)=\frac{1/4}{5/12}=3/5$.

Hint: typically the denominator will include the term in the numerator in this kind of questions, since the numerator is always a subset of the term in the denominator (I mean the sets).