Question

An urn contains 4 white and 3 red balls. Let x be the number of red balls in a random draw of three balls. Find the mean and variance of x

Answer 1

Answer:

$Mean\,of\,x=\frac{9}{7}\:\:and\:\:variance\,of\,x=\frac{36}{49}$

Step-by-step explanation:

Given : an URN contains 4 white & 3 Red balls

To find: Mean & Variance of X

Let X be the number of Red balls in a random draw of three balls

⇒ X takes value of 0 , 1 , 2 , 3

Total No of balls in URn = 3 + 4 = 7

$Probability\:of\;get\;a\:Red\:ball\,=\,\frac{3}{7}$

$Probability\:of\;get\;a\:White\:ball\,=\,\frac{4}{7}$

So,

$P(X=0)=P(no\,red\,ball)=\frac{4}{7}\times\frac{4}{7}\times\frac{4}{7}=\frac{64}{343}$

$P(X=1)=P(1\,red\,ball\,and\,2\,white\,ball)=\frac{3}{7}\times\frac{4}{7}\times\frac{4}{7}+\frac{4}{7}\times\frac{3}{7}\times\frac{4}{7}+\frac{4}{7}\times\frac{4}{7}\times\frac{3}{7}=\frac{144}{343}$

$P(X=2)=P(2\,red\,ball\,and\,1\,white\,ball)=\frac{3}{7}\times\frac{3}{7}\times\frac{4}{7}+\frac{3}{7}\times\frac{4}{7}\times\frac{3}{7}+\frac{4}{7}\times\frac{3}{7}\times\frac{3}{7}=\frac{108}{343}$

$P(X=3)=P(3\,red\,ball)=\frac{3}{7}\times\frac{3}{7}\times\frac{3}{7}=\frac{27}{343}$

Table we get is attached,

Mean of x,

$E(X)=\sum_{i=1}^{n}x_ip_i=0\times\frac{64}{343}+1\times\frac{144}{343}+2\times\frac{108}{343}+3\times\frac{27}{343}\\\\E(X)=\frac{441}{343}=\frac{9}{7}$

formula used for variance of X is given by,

var(X) = E(X²) - [E(X)]²

$E(X^2)=\sum_{i=1}^{n}x_i^2p_i=0^2\times\frac{64}{343}+1^2\times\frac{144}{343}+2^2\times\frac{108}{343}+3^2\times\frac{27}{343}\\\\E(X^2)=\frac{819}{343}$

Thus, $Var(X)=\frac{819}{343}-(\frac{441}{343})^2$

$Var(X)=\frac{86436}{117649}=\frac{36}{49}$

Therefore, $Mean\,of\,x=\frac{9}{7}\:\:and\:\:variance\,of\,x=\frac{36}{49}$