Two identical urns contain balls. One of the urns has 6 red balls and 3 blue balls. The other urn has 5 red balls and 8 blue balls. An urn is chosen at random and a ball is drawn at random from this urn. If the ball turns out to be blue, what is the probability that this is the urn with 8 blue balls?
— My first thought was that the answer to this question would be as simple as 8 blue ball instances can be drawn from the urn with a total sample space of 11 blue balls. This previously led me to believe that the probability is 8/11. However, I feel the answer is more complicated than this. Should I use bayes-theorem to compute the probability of this question? If so, what would the solution be? Thanks in advance.
Best Answer
We have: Urn $X$ is 6 red: 3 blue, Urn $Y$ is 5 red: 8 blue.
Which is to say: $\mathsf P(B\mid X) = \frac 3 9, \mathsf P(B\mid Y)= \frac 8 {13}$ and if the urns are chosen without bias then: $\mathsf P(X)=\mathsf P(Y)$
Thus $\mathsf P(Y\mid B) = \dfrac{\mathsf P(B\mid Y)\mathsf P(Y)}{\mathsf P(B\mid Y)\mathsf P(Y)+\mathsf P(B\mid X)\mathsf P(X)} = \dfrac{8/13}{8/13+3/9} = \dfrac{24}{37}$
The probability that the blue ball selected came from the urn with eight blue balls is: $24/37$ and not $8/11$ as naively suggested.
tl;dr Because it is the urns themselves that have an equal probability of being selected, but not the individual balls, therefore "yes", you do have to use Bayes' Theorem.