# Permutations and Conditional Probability

I was going through the problem that I have mentioned below. Included is the proposed solution, however, I want to be crystal clear about the reasoning behind the expressions used to compute the conditional probabilities:

An urn has 'r' red balls and 'w' white balls that are removed one at a time randomly. Let $$R_i$$ be the event where the $$i^\text{th}$$ ball is red. Compute:

(1) $$P(R_i)$$
(2) $$P(R_5 \mid R_3 )$$

(1) All balls are distinguishable (with ordering), thus giving the number of permutations of 'i' balls:

$$= \frac{(r+w)!}{(r+w-i)!}$$

Outcomes in $$R_i =$$ ($$r$$ ways to chose $$i^\text{th}$$ ball) $$\cdot \left( \frac{(r + w - 1)!}{(r + w - 1 - (i -1))!} \right) = r \cdot \frac{(r+w-1)!}{(r+w-i)!}$$

[Shouldn't the numerator be $$(r+w)!$$.Why are already we assuming one ball has been taken? Is it because we are simply calculating the permutations of the remaining balls? This seems pretty obvious because otherwise the probability will become $$r$$ which is obviously not true, but it still doesn't explain my question.]

$$\therefore P(R_i) = \dfrac{ \frac{(r + w -1)!}{(r + w -i)!} }{\frac{(r+w)!}{(r+w-i)!}}$$

$$\therefore P(R_i) = \dfrac{r(r+w-1)}{(r+w)!} = \dfrac{r}{r+w}$$

(2) In this part the calculation of $$P(R_5 \cap R_3)$$ seems to follow the same principle.

Why are we dividing by the number of balls left?

$^{(r+w)}{\rm P}_{i} = \frac{(r+w)!}{(r+w-i)!}$ is the count of ways to extract $i$ distinct balls (line them up in order of extraction).

$^{(r+w)}{\rm P}_{i} = {^{(r+w)}{\rm C}_{i}}\times i!$

Shouldn't the numerator be $(r+w)!$ since $i=1$ is a possibility. Why are already we assuming one ball has been taken? Is it because we are simply calculating the permutations of the remaining balls?]

Yes.   $^{(r+w-1)}{\rm P}_{(i-1)} = \frac{(r+w-1)!}{(r+w-1-(i-1))!}$ is the count of ways to extract $i-1$ distinct balls before extracting a particular one in the $i$-th place.   When that $i$-th ball is given, there are only $r+w-1$ from which to select the others.

Why does this expression have $(r+w)$, the the total initial number in denominator's numerator?]

$$\require{cancel}\frac{(r+w-1)!}{(r+w)!} = \frac{\color{silver}{\cancel{\color{black}{(r+w-1)(r+w-2)\cdots 1}}}}{(r+w)\color{silver}{\cancel{\color{black}{(r+w-1)(r+w-2)\cdots 1}}}} \\ \qquad\qquad =\frac{1}{r+w}$$