A bag contains $4$ red balls and $6$ black balls. A person randomly takes out one ball from the bag, sees it's color and put the ball back into the bag alongwith two additional balls of the color he saw. Now he again takes out a ball randomly. Find the probability that the ball which is taken out is red.
I know this is a classic conditional probability problem but I have some unusual approach (actually it's of my ex-teacher) for this problem.
We will assume that there is $4L$ red paint and $6L$ black paint in a bucket. The person took out $1L$ of paint. Since the concentration of paint is uniform, he actually took out $\frac{4}{10}L$ of red paint and $\frac{6}{10}L$ of black paint.
Now he will see $\frac{4}{10}L$ of red paint and $\frac{6}{10}L$ of black paint.
This simply means that he will put an additional $2×\frac{4}{10}L$ of red paint and an additional $2×\frac{6}{10}L$ of black paint into the bucket.
Now the bucket contains $4+2×\frac{4}{10}L=X$ of red paint and $6+2×\frac{6}{10}L=Y$ of black paint.
Now the probability of taking out a red ball from the bag is simply $\frac{X}{X+Y}$
Why does this approach works$?$ This approach skips all the tedious and hard calculations and concepts of Baye's theorem and calculates the probability through elementary ways. Why don't they teach this thing in schools$?$
Any discussion is greatly appreciated.