Urn A contains 6 red and 4 black balls. Urn B contains 4 red and 6 black balls. One ball is drawn at random from Urn A and placed in Urn B. Then, one ball is drawn at random from Urn B and placed in Urn A. Now, If one ball is drawn from Urn A, find the probability that it is red. (Cengage JEE Adv Algebra, 3e)
The question is straight forward. The general way of doing it would be to consider each possibility in which the last ball drawn is red and adding all of them up.
As per the solution in the book:
Case 0: Red -> Red -> Red
Case 1: Black -> Red -> Red
Case 2: Red -> Black -> Red
Case 3: Black -> Black -> Red
Find the probabilities of each individual cases, and add them up to get $\frac{32}{55}$.
But there is another way of solving this question. Its not easier, per se, but much more interesting.
Instead of thinking that the balls will either be of one discrete color, think of the color as “non-discrete” property, i.e, a ball can have a certain % of red and remaining % of black.
Urn A -> 6 red + 4 black
If I were to choose any random ball from Urn A, I could imagine that ball to have 0.6 red and 0.4 black property (individual probabilities of choosing red and black balls respectively). In other words, as long as you donot observe the color of the picked ball, it will be in superposition of both red and black.( Note that this does not mean that the actual ball is both red and black. It just a simple way of adding above cases(possibilities) in just one step )
Forget about the balls being discrete. Think only about the color. When I pick any random ball from Urn A and put it in Urn B, the probabilities(of selecting colored balls) are hampered in both Urns. A ball is removed from Urn A, so red and black properties decreases. A ball is added in Urn B, so red and black properties increases.
Urn A -> (6 - 0.6)red + (4 - 0.4)black = 5.4 red balls + 3.6 black balls = 9 balls
Urn B -> (4 + 0.6)red + (6 + 0.4)black = 4.6 red balls + 6.4 black balls = 11 balls
Now, we need to do the same step again; this time, from Urn B to Urn A.
If I were to now pick one random ball from Urn B, it will have (4.6/11 red + 6.4/11 black).
Again, Picking this ball and putting it in Urn A will hamper the probabilities of both urns.
Urn A -> (5.4 + 4.6/11)red balls + (3.6 + 6.4/11)black balls = 10 balls
Urn B -> (4.6 - 4.6/11)red balls + (6.4 - 6.4/11)black balls = 10 balls
Now, it is the last part of the question. If we pick one ball from Urn A, find the probability that it is red.
$$\implies P(red)=\frac{5.4 + \frac{4.6}{11}}{10}$$ $$\implies P(red)=\frac{64}{110} = \frac{32}{55}$$
While the solution might seem a bit long and confusing, if you use diagrams, it will become much shorter(than the first solution).