Q.7.41 - A First Course in Probability

Question

A pond contains $100$ fish, of which $30$ are carp. If $20$ fish are caught, what are the mean and variance of the number of carp among the $20$? What assumptions are you making?

Step-by-Step Solution

Answer

The assumptions made are $X$ follows Hyper geometric distribution with$m=30, n=20, N=100$. That is, $X~HG(N=100,m=30,n=20)$.

Step 1: Given Information

The total number of fish caught is, $N=100$

The number of carp is, $m=30$

The number of fish caught is, $n=20$

The mean and variance of the number of carp among the $20=?$

Step 2: Explanation

From the information, observe that a pond contains 100 fish, of which 30 are carp.

There are 20 fishes are caught from the pond.

Consider $X$ is the random variable that represents the number of carp caught of the 20 fishes caught.

Assume that the catch of 20 fishes is equally likely.

The random variable $X$ follows Hypergeometric distribution with parameters with parameters $N=100, m=30$and $n=20$

Calculate mean number of carp,

$E\left(X\right)=\frac{nm}{N}$

$=\frac{20\times 30}{100}$

$=6$

Step 3: Explanation

Calculate the variance of the number of carp.

$V\left(X\right)=\frac{mn}{N}\left[\frac{(n-1)(m-1)}{N-1}+1-\frac{mn}{N}\right]$

$=\frac{30\times 20}{100}\left[\frac{(20-1)(30-1)}{100-1}+1-\frac{30\times 20}{100}\right]$

$=\frac{600}{100}\left[\frac{19\times 29}{99}+1-\frac{600}{100}\right]$

$=\frac{600}{100}\left[\frac{551}{99}+1-\frac{600}{100}\right]$

$=\frac{600}{100}\left[\frac{56}{99}\right]$

$=\frac{112}{33}$

Step 4: Final Answer

The assumptions made are $X$ follows Hyper geometric distribution with $m=30, n=20, N=100$. That is, $X ~ H G(N=100, m=30, n=20).$