Hypergeometric Distribution - Order of Draws

Order of Draws

The probability of drawing any sequence of white and black marbles (the hypergeometric distribution) depends only on the number of white and black marbles, not on the order in which they appear; i.e., it is an exchangeable distribution. As a result, the probability of drawing a white marble in the draw is

This can be shown by induction. First, it is certainly true for the first draw that:

.

Also, we can show that by writing:

\begin{align} P(W_{n+1}) & = {\sum_{k=0}^n}P(W_{n+1}|k)f(k;N,m,n)\\ & = {\sum_{k=0}^n}\frac{m-k}{N-n}f(k;N,m,n) \\ & = {\sum_{k=0}^n}\frac{m-k}{N-n}\frac{\binom mk \binom {N-m} {n-k}}{\binom Nn} \\ & = \frac{1}{(N-n)\binom Nn} \left \{ m\sum_{k=0}^n \binom mk \binom {N-m} {n-k} - \sum_{k=0}^n k\binom mk \binom {N-m} {n-k}\right \} \\ & = \frac{1}{(N-n)\binom Nn}\left\{ m\binom Nn - \sum_{k=1}^n k\frac{m}{k} \binom {m-1}{k-1} \binom {N-m} {n-k}\right \} \\ & = \frac{m}{(N-n)\binom Nn}\left\{ \binom Nn - \sum_{k=1}^n \binom {m-1}{k-1} \binom {N-1-(m-1)} {n-1-(k-1)}\right \} \\ & = \frac{m}{(N-n)\binom Nn}\left\{ \binom Nn - \binom {N-1}{n-1}\right \} \\ & = \frac{m}{(N-n)\binom Nn}\left\{ \binom Nn - \frac{n}{N}\binom Nn\right \} \\ & = \frac{m}{(N-n)}\left\{ 1 - \frac{n}{N} \right\} = \frac{m}{N} \end{align} ,

which makes it true for every draw.

A simpler proof than the one above is the following:

By symmetry each of the marbles has the same chance to be drawn in the draw. In addition, according to the sum rule, the chance of drawing a white marble in the draw can be calculated by summing the chances of each individual white marble being drawn in the . These two observations imply that if for example the number of white marbles at the outset is 3 times the number of black marbles, then also the chance of a white marble being drawn in the draw is 3 times as big as a black marble being drawn in the draw. In the general case we have white marbles and black marbles at the outset. So

.

Since in the draw either a white or a black marble needs to be drawn, we also know that

.

Combining these two equations immediately yields

.

Read more about this topic:  Hypergeometric Distribution

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