Q. 4.29 - A First Course in Probability

Question

For a hypergeometric random variable, determine 

$P\{X=k+1\}/P\{X=k\}$

Step-by-Step Solution

Answer

$\frac{(K-k)(n-k)}{(k+1)(N-K-n+k+1)}$

Step 1 Given information

For Hypergeometric random variable with parameters $N,K,n$ we have that

$P(X=k)=\frac{\left(\begin{array}{c}K\\ k\end{array}\right)·\left(\begin{array}{c}N-K\\ n-k\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$

Step 2 Calculation

So we have that 

$\frac{P(X=k+1)}{P(X=k)}=\frac{\frac{\left(\begin{array}{c}K\\ k+1\end{array}\right)·\left(\begin{array}{c}N-K\\ n-(k+1)\end{array}\right)}{\left(\begin{array}{l}N\\ n\end{array}\right)}}{\frac{\left(\begin{array}{c}K\\ k\end{array}\right)·\left(\begin{array}{c}N-K\\ n-k\end{array}\right)}{\left(\begin{array}{l}N\\ n\end{array}\right)}}=\frac{\left(\begin{array}{c}K\\ k+1\end{array}\right)·\left(\begin{array}{c}N-K\\ n-(k+1)\end{array}\right)}{\left(\begin{array}{c}K\\ k\end{array}\right)·\left(\begin{array}{c}N-K\\ n-k\end{array}\right)}$

Step 3 Continue Calculation

When we write out these binomial coefficients, we get that the expression above is equal to 

$\frac{\frac{K!}{(k+1)!(K-k-1)!}·\frac{(N-K)!}{(n-k-1)!(N-K-n+k+1)!}}{\frac{K!}{k!(K-k)!}·\frac{(N-K)!}{(n-k)!(N-K-n+k)!}}$

Step 4 Final answer

If we cancel out everything we can we left with.

$\frac{(K-k)(n-k)}{(k+1)(N-K-n+k+1)}$