## Hyper-geometric distribution:

### Suppose a population consists of N items, k of which are successes. A random sample drawn from that population consists of n items, x of which are successes.

**Then the hyper-geometric probability is:**

h(x, N, n, k) = [ ^{k}C_{x}][^{(n-k)}C_{(n-k)}]/[^{N}C_{n}]

Where,

N -Number of items in the population.

k: Number of items in the population that are classified as successes.

n: Number of items in the sample.

x: Number of items in the sample that are classified as successes.

**10 cards are selected randomly from a deck of cards without replacements. What is the probability of getting exactly 5 black cards?**

**Solution:**

*N = 52, k = 26; since there are 26 black cards in a deck.**
*- n = 10; since we randomly select 10 cards from the deck.
- x = 5; since 5 of the cards we select are black.
- h(x; N, n, k) = [
^{k}C_{x}][^{(n-k)}C_{(n-k)}]/[^{N}C_{n}]
- h(5; 52, 10, 26) = [
^{26}C_{5}*^{26}C_{5}]/[^{52}C_{10}]
*h(5; 52, 10, 26) = [ 65780 * 65780] / [ 15820024220 ] = 0.2735*

**There are 20 black marbles and 20 white marbles out of which 10 marbles are being chosen. Find the probability that there are 4 white marbles in them.**

**Solution:**

*N = 40, k = 20; since there are 20 white marbles.**
*- n = 10; randomly select 10 cards from the deck.
- x = 4; since 4 of the white marbles .
- h(x; N, n, k) = [
^{k}C_{x}][^{(n-k)}C_{(n-k)}]/[^{N}C_{n}]
- h(4; 40, 10, 20) = [
^{20}C_{4}*^{20}C_{6}]/[^{40}C_{10}]
*h(4; 40, 10, 20) = [ 4845 * 38760] / [ 847660528 ] = 0.2215 *

**Out of 150 aspirants qualifying an exam, 25 were drawn randomly. If 30 out of 150 qualified aspirants are female, then find the probability that 8 out of 25 chosen are females.**

**Solution:**

*N = 150, k = 30;**
*- n = 25
- x = 8;
- h(x; N, n, k) = [
^{k}C_{x}][^{(N-k)}C_{(n-x)}]/[^{N}C_{n}]
- h(8; 150, 25, 30) = [
^{30}C_{8}*^{120}C_{17}]/[^{150}C_{25}]
*h(8; 150, 25, 30) = 0.0568*

**There are 66 bulbs in a house out which 33 are defective. If 22 bulbs are picked randomly, find the probability that at least one is defective.**

**Solution:**

*h(x; N, n, k) = [*^{k}C_{x}][^{(N-k)}C_{(n-x)}]/[^{N}C_{n}]*
*- P(X=0) = [
^{3}C_{0}*^{3}C_{2}]/[^{6}C_{2}]
- = 0.2
- = 0.2
- Probability of at least one defective bulb = 1 – p(X=0) = 1 – 0.2
*= 0.8*

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