There are two bags, $A$ and $B$ having $2$ white balls and $1$ black ball in Bag $A$ and $1$ white ball and $3$ black balls in bag $B$. A bag is chosen at random and two balls are drawn. The balls were $1$ white and $1$ black ball. What is the probability that is from bag $B$?

I have solved it visually using a tree structure, but how I express it in mathematical expression.

My Solution


closed as off-topic by Saad, Remy, TheSimpliFire, A. Goodier, mlc Mar 4 '18 at 11:50

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  • $\begingroup$ Can't understand how to do it! I have tried with 1 ball from bag B $\endgroup$ – Siddharth Murari Mar 4 '18 at 8:28

Let $A,B$ be the complementary events of selecting the relevant bag.   Let $E$ be the evidence: the event that two balls drawn are white and black.   You seek the (conditional)probability that bag B was choosen given this evidence.

This is an application of Bayes' rule and the Law of Total Probability.

$$\mathsf P(B\mid E)~{=\dfrac{\mathsf P(E\mid B)~\mathsf P(B)}{\mathsf P(E\mid A)~\mathsf P(A)+\mathsf P(E\mid B)~\mathsf P(B)}}$$

You can evaluate these probabilities from your tree.


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