Introduction to Hypergeometric Distribution
The hypergeometric probability distribution models scenarios where items are selected without replacement. It applies when sampling from a finite population with two distinct groups (successes and failures).
The combination formula is fundamental:
\[
\binom{M}{m} = \frac{M!}{m!(M-m)!}
\]
where \( M! = 1 \times 2 \times 3 \times \cdots \times (M-1) \times M \). This represents the number of ways to choose \( m \) items from \( M \) distinct items without regard to order.
Solved Hypergeometric Probability Examples
Example 1: Selecting Colored Balls
Four balls are randomly selected from a box containing 5 red and 3 white balls.
- Find the probability of selecting equal numbers of red and white balls.
- Find the probability of selecting at least 2 red balls.
Solution
Given: \( N = 8 \), \( R = 5 \) (red), \( N-R = 3 \) (white), \( n = 4 \).
Part (a): We need \( x = 2 \) red balls (and thus 2 white).
\[
P(X = 2) = \frac{\binom{5}{2} \binom{3}{2}}{\binom{8}{4}} = \frac{10 \times 3}{70} = \frac{3}{7}
\]
Part (b): "At least 2 red" means \( X \ge 2 \):
\[
\begin{aligned}
P(X \ge 2) &= P(X=2) + P(X=3) + P(X=4) \\
&= \frac{\binom{5}{2}\binom{3}{2}}{\binom{8}{4}} + \frac{\binom{5}{3}\binom{3}{1}}{\binom{8}{4}} + \frac{\binom{5}{4}\binom{3}{0}}{\binom{8}{4}} \\
&= \frac{3}{7} + \frac{3}{7} + \frac{1}{14} = \frac{13}{14}
\end{aligned}
\]
Example 2: Quality Control Inspection
An inspector selects 4 tools from 15, where 2 are defective.
- Probability of exactly one defective among the 4.
- Probability of no defective tools.
- How many non-defective tools must be added so that the probability of zero defectives becomes 0.70?
Solution
Given: \( N = 15 \), defective = 2 (successes), non-defective = 13, \( n = 4 \).
Part (a): Let \( X \) be the number of defective tools. We want \( P(X = 1) \):
\[
P(X = 1) = \frac{\binom{2}{1} \binom{13}{3}}{\binom{15}{4}} = \frac{2 \times 286}{1365} \approx 0.41905
\]
Part (b): No defectives means 4 non-defective:
\[
P(X = 4) = \frac{\binom{13}{4} \binom{2}{0}}{\binom{15}{4}} = \frac{715 \times 1}{1365} \approx 0.52381
\]
Part (c): Let \( y \) = additional non-defective tools. New totals: \( N = 15 + y \), \( R = 13 + y \) (non-defective). We want:
\[
P(\text{4 non-defective}) = \frac{\binom{13+y}{4} \binom{2}{0}}{\binom{15+y}{4}} = 0.70
\]
Simplifying the combination ratio:
\[
\frac{(13+y)!}{4!(9+y)!} \times \frac{4!(11+y)!}{(15+y)!} = 0.70
\]
\[
\frac{(10+y)(11+y)}{(14+y)(15+y)} = 0.70
\]
Cross-multiplying and simplifying yields:
\[
0.3y^2 + 0.7y - 37 = 0
\]
The positive solution is \( y = 10 \). Thus, add 10 non-defective tools.
Example 3: Committee Selection
A 6-person committee is randomly selected from 4 men and 8 women.
- Probability of equal gender representation.
- Probability of at most 3 men.
Solution
Given: \( N = 12 \), \( R = 4 \) (men), \( N-R = 8 \) (women), \( n = 6 \).
Part (a): Equal representation means 3 men and 3 women:
\[
P(X = 3) = \frac{\binom{4}{3} \binom{8}{3}}{\binom{12}{6}} = \frac{4 \times 56}{924} = \frac{8}{33}
\]
Part (b): "At most 3 men" includes \( X = 0,1,2,3 \):
\[
\begin{aligned}
P(X \le 3) &= \sum_{x=0}^{3} \frac{\binom{4}{x} \binom{8}{6-x}}{\binom{12}{6}} \\
&= \frac{\binom{4}{0}\binom{8}{6} + \binom{4}{1}\binom{8}{5} + \binom{4}{2}\binom{8}{4} + \binom{4}{3}\binom{8}{3}}{\binom{12}{6}} \\
&= \frac{1 \cdot 28 + 4 \cdot 56 + 6 \cdot 70 + 4 \cdot 56}{924} = \frac{32}{33}
\end{aligned}
\]
Example 4: Multicolor Selection
Six balls are randomly selected from 6 red, 4 blue, and 5 white balls. Find the probability of selecting 3 red, 2 blue, and 1 white ball.
Solution
This is a multivariate hypergeometric scenario. Total balls: \( N = 15 \), sample size \( n = 6 \).
\[
P(\text{3R, 2B, 1W}) = \frac{\binom{6}{3} \binom{4}{2} \binom{5}{1}}{\binom{15}{6}} = \frac{20 \times 6 \times 5}{5005} = \frac{120}{1001}
\]
Example 5: Even Numbers Lottery
Seven numbers are randomly selected from integers 1–49. For what number of even numbers \( x \) is the probability \( P(X=x) \) maximized?
Solution
There are 24 even numbers (2,4,…,48) and 25 odd numbers (1,3,…,49). We sample \( n = 7 \). The probability is:
\[
P(X = x) = \frac{\binom{24}{x} \binom{25}{7-x}}{\binom{49}{7}}
\]
Computing for \( x = 0 \) to \( 7 \):
- \( P(X=0) = \frac{\binom{24}{0} \binom{25}{7}}{\binom{49}{7}} \approx 0.00560 \)
- \( P(X=1) = \frac{\binom{24}{1} \binom{25}{6}}{\binom{49}{7}} \approx 0.04948 \)
- \( P(X=2) = \frac{\binom{24}{2} \binom{25}{5}}{\binom{49}{7}} \approx 0.17071 \)
- \( P(X=3) = \frac{\binom{24}{3} \binom{25}{4}}{\binom{49}{7}} \approx 0.29806 \) (maximum)
- \( P(X=4) = \frac{\binom{24}{4} \binom{25}{3}}{\binom{49}{7}} \approx 0.28451 \)
- \( P(X=5) = \frac{\binom{24}{5} \binom{25}{2}}{\binom{49}{7}} \approx 0.14844 \)
- \( P(X=6) = \frac{\binom{24}{6} \binom{25}{1}}{\binom{49}{7}} \approx 0.03917 \)
- \( P(X=7) = \frac{\binom{24}{7} \binom{25}{0}}{\binom{49}{7}} \approx 0.00403 \)
The probability is highest when \( x = 3 \) even numbers.
