How to derive the following combinatorial result? (Conditional Probability Problem)

Question

The problem is an example from the book Introduction to The Theory of Statistics by Mood. It´s the example 28, page 39:

"Suppose an urn contains $M$ balls of which K are black and $M-K$ are white. A sample of size n is drawn. Fiund the probability that the jth ball drawn is black given that the sample contains k black balls. (We intuitively expect the answer to be k/n.) We have to consider sampling (i) with replacement and (ii) without replacement. [...]

SOLUTION

Let $A_k$ denote the event that the sample contains exactly $k$ black balls and $B_j$ denote the event that the $jth$ ball drawn is black. We seek $P[B_j|A_k]$

For case (ii),

$P[A_k] = \frac{\binom{K}{k} \binom{M-K}{n-k}}{ \binom{M}{n}}$ and $P[A_k|B_j] = \frac{\binom{K-1}{k-1} \binom{M -K}{n-k}}{ \binom{M-1}{n-1}}$

$$P[B_j] = \sum_{i=0}^{j-1} P[B_j|C_i]P[C_i]$$, where $C_i$ denotes the event of exactly $i$ black balls in the first $j-1$ draws. Note that

$$P[C_i] = \frac{ \binom{K}{i} \binom{M-K}{j-1-i} }{\binom{M}{j-1}} $$

and $$ P[B_j|C_i] = \frac{K-i}{M-j+1} $$ and so

$$ P[B_j] = \sum_{i=0}^{j-1}\frac{K-i}{M-j+1}\frac{\binom{K}{i} \binom{M-K}{j-1-i} }{\binom{M}{j-1}}=\frac{K}{M} $$

Finally $$P[B_j|A_k] = \frac{P[A_k|B_j]P[B_j]}{P[A_k]}= \frac{ \frac{ \binom{K-1}{k-1} \binom{M-K}{n-k} }{ \binom{M-1}{n-1} } \frac{K}{M} }{\frac{\binom{K}{k} \binom{M-K}{n-k}}{ \binom{M}{n}}} = \frac{k}{n}$$"

I understand the problem and the answer, I just cannot understand one of the steps, which is the next sum of binomial coefficients: $$ P[B_j] = \sum_{i=0}^{j-1}\frac{K-i}{M-j+1}\frac{\binom{K}{i} \binom{M-K}{j-1-i} }{\binom{M}{j-1}}=\frac{K}{M} $$

I have tried but I have not got the answer yet, could anyone provide a hint or an explination of how to get that answer?

Thanks in advance!

Answer 1

I think there is a typo, specifically it should be $M-j+1$ instead of $M-j+i$. Indeed, the equality as written fails at $i=0$ and $j=1$.

After making this change, the problem follows by simple manipulation :

$P[B_j] = \sum_{i=0}^{j-1} \dfrac{K}{M} \dfrac{{K-1 \choose i} {M-K \choose j-1-i}}{{M-1 \choose j-1}} = \dfrac{K}{M}$

since the sum of the numerator counts the number of ways of choosing arbitrary $j-1$ people out of a crowd of $K-1$ girls and $M-K$ boys.

Answer 2

Let $0\le K\le M$ and $1\le j\le M$. Consider a well-shuffled deck of $M$ cards, $K$ of which are black, and let $B_j$ be the event that the $j^\text{th}$ card in the deck is black. I will obtain two different expressions for the probability $P(B_j)$.

First, it is clear that $$P(B_j)=\frac KM\tag1$$ since $P(B_1)=P(B_2)=\cdots=P(B_M)$ and $\sum_{j=1}^MP(B_j)=K$.

Now let $A_{i,j}$ be the event that exactly $i$ of the first $j-1$ cards are black. Then $$P(B_j)=\sum_{i=0}^{j-1}P(B_j|A_{i,j})P(A_{i,j})=\sum_{i=0}^{j-1}\frac{K-i}{M-j+1}\cdot\frac{\binom Ki\binom{M-K}{j-1-i}}{\binom M{j-1}}.\tag2$$ Combining $(1)$ and $(2)$ we get $$\sum_{i=0}^{j-1}\frac{K-i}{M-j+1}\cdot\frac{\binom Ki\binom{M-K}{j-1-i}}{\binom M{j-1}}=P(B_j)=\frac KM.$$

Answer 3

After going through answer provided by bof I feel that the final answer can be calculated by considering the situation in a different way.

Let us assume that we draw all the $M$ cards and lay them in a row. At the j-th draw you want a black card.

Number of ways to arrange the cards is $M!$.

Number of ways to choose one black card for the j-th position is $\binom {K}{1}$ ways.

Remaining $(M-1)$ cards can be arranged in $(M-1)!$ ways.

So the required probability= $\frac{\binom {K}{1}\times (M-1)!}{M!}=\frac{K}{M}$

Alternatively, each of the $M$ positions can have a black card with probability $\frac{K}{M}$