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 $\dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6}$?


1 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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.