There are n total balls in the urn containing red and blue colors with unknown ratios. k red balls can be anywhere from 0 to n being equally likely and the remaining n–k will be blue balls. What is the probability of k red balls in the urn if the first ball drawn is red? What is the probability of k red balls in the urn if the first ball drawn is red without replacement and the second ball drawn is blue?
I know that Bayes's rule will be applied to find out the probability of red balls present in the given. I think P(k red balls) = 1/(n+1) and P(n–k blue balls) = n/n+1
Further, Bayes's formula will be applied so that
P(k red balls|ball drawn is red) = {P(ball drawn is red | k red balls)*P(k red balls)}/P (ball drawn is red)
But I am stuck as to how to use this to find the answer.
Best Answer
The probability of $k$ red balls and the first ball drawn being red is $\frac1{n+1}$ (for the equally likely initial distributions) times $\frac kn$. Total probability (a more intuitive form of the Price–Bayes formula) then gives the answer as $$\frac{(1/(n+1))(k/n)}{(1/(n+1))\sum_{k=0}^n(k/n)}=\frac{k}{\sum_{k=0}^nk}=\frac{2k}{n(n+1)}$$ Likewise the probability of $k$ red balls, the first draw red and the second draw blue is $\frac1{n+1}\frac kn\frac{n-k}{n-1}$, so the second part's answer is $$\frac{(1/(n+1))(k/n)((n-k)/(n-1))}{(1/(n+1))\sum_{k=0}^n(k/n)((n-k)/(n-1))}=\frac{k(n-k)}{\sum_{k=0}^nk(n-k)}=\frac{6k(n-k)}{(n-1)n(n+1)}$$