Solved: Suppose that X has a hypergeometric distribution

Step 1 of 4

(a) Suppose that \(X\) has a hypergeometric distribution with \(N=100\), \(n=4\) and \(K=20\),

we are asked to find the probability \(P(X=1)\).

A set of \(N\) object contains \(K\) objects classified as successes \(N-K\) objects classified as failures.

A sample of size \(n\) objects is randomly (without replacement) from the \(N\) objects where \(K \leq N\) and \(n \leq N\).

The random variable \(X\) that equals the number of successes in the sample is a hypergeometric random variable and

\(f(x)=\frac{\left(\begin{array}{l} k \\ x \end{array}\right)\left(\begin{array}{c} N-K \\ n-x \end{array}\right)}{\left(\begin{array}{l} N \\ n \end{array}\right)} \quad x=\max \{0, n+K-N\} \text { to } \min \{k, n\}\) ………… (1)

Hence we can substitute the given values into equation (1) with \(x=6\).

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

\(f(1)=P(X=1)=\frac{\left(\begin{array}{c} 20 \\ 1 \end{array}\right)\left(\begin{array}{c} 100-20 \\ 4-1 \end{array}\right)}{\left(\begin{array}{c} 100 \\ 4 \end{array}\right)}=0.419\)

Hence the probability \(P(X=1)\) is 0.419