Question

# An urn contains 3 white and 6 red balls. Four balls are drawn one by one with replacement form the urn. Find the probability distribution of the number of red balls drawn. Also, find the mean and variance of the distribution.

Hint: Let the number of red balls drawn from the urn with replacement be represented by the random variable $X$. Find the probability for the different values of $X$, that is 0, 1, 2, 3 and 4.
Find the mean using the direct formula $np$ and the variance $npq$, where $p$ is the probability of success and $q$ is the probability of failure.

Complete step by step solution:
Let us assume the number of red balls drawn from the urn represented by the random variable $X$. Since the total of 4 balls have been drawn from the urn, therefore the possible values of $X$ are 0,1,2,3 and 4.
Since there is replacement after every ball is drawn, there are always 3 white balls and 6 red balls in the urn.
The probability of drawing a white ball from the urn is, therefore, $\dfrac{3}{9}$ since there are 3 white balls.
$P\left( W \right) = \dfrac{1}{3}$
The probability of drawing a red ball from the urn is, therefore, $\dfrac{6}{9}$ since there are 6 red balls.
$P\left( R \right) = \dfrac{2}{3}$
The condition $X = 0$ specify that no red ball must be drawn from the urn, and can be found by multiplying the probability of getting a white ball four times to the number of ways in which the white ball can be picked.
$P\left( {X = 0} \right){ = ^4}{C_0}{\left( {\dfrac{1}{3}} \right)^4}{\left( {\dfrac{2}{3}} \right)^0}$

Similarly, when we have $X = 1$, it implies that one red ball is drawn from the urn and can be calculated as,
$P\left( {X = 1} \right){ = ^4}{C_1}{\left( {\dfrac{1}{3}} \right)^3}{\left( {\dfrac{2}{3}} \right)^1}$
Now, find the probability when $X = 2$
$P\left( {X = 2} \right){ = ^4}{C_2}{\left( {\dfrac{1}{3}} \right)^2}{\left( {\dfrac{2}{3}} \right)^2}$
We will also find the probability when three red balls are drawn when $X = 3$
Hence, $P\left( {X = 3} \right){ = ^4}{C_3}{\left( {\dfrac{1}{3}} \right)^1}{\left( {\dfrac{2}{3}} \right)^3}$
Also, there is a possible condition when all red balls are drawn and no white ball is drawn.
Then, we have $X = 4$ and probability is given as,
$P\left( {X = 4} \right){ = ^4}{C_4}{\left( {\dfrac{1}{3}} \right)^0}{\left( {\dfrac{2}{3}} \right)^4}$
We can represent all the cases in tabular format.
Hence, the probability distribution of the number of red balls drawn is:

 $X$ $P\left( X \right)$ 0 $^4{C_0}{\left( {\dfrac{1}{3}} \right)^4}{\left( {\dfrac{2}{3}} \right)^0}$ 1 $^4{C_1}{\left( {\dfrac{1}{3}} \right)^3}{\left( {\dfrac{2}{3}} \right)^1}$ 2 $^4{C_2}{\left( {\dfrac{1}{3}} \right)^2}{\left( {\dfrac{2}{3}} \right)^2}$ 3 $^4{C_3}{\left( {\dfrac{1}{3}} \right)^1}{\left( {\dfrac{2}{3}} \right)^3}$

Since there are only 2 possible outcomes, this is also known as a binomial distribution.
Now, we know that the mean can be calculated as $np$ in case of the binomial distribution.
Here, $n = 4$ as four balls are getting drawn and $p = \dfrac{2}{3}$ is the probability of drawing the red ball.
Then, mean $\mu$ is
$\mu = 4 \times \dfrac{2}{3} \\ \Rightarrow \mu = \dfrac{8}{3} \\$
Next, the variance is given by $npq$, where $p$ is the probability of not getting a red ball from the urn.
Therefore, we have variance ${\sigma ^2}$ as ,
${\sigma ^2} = 4 \times \dfrac{2}{3} \times \left( {1 - \dfrac{2}{3}} \right) \\ \Rightarrow {\sigma ^2} = 4 \times \dfrac{2}{3} \times \dfrac{1}{3} \\ \Rightarrow {\sigma ^2} = \dfrac{8}{9} \\$

Note: Binomial distributions can only be possible when there are only possible outcomes where each of the trials is independent of the other and the probability of one should be equal to the probability of not getting the other. Or, we can say that the sum of probability of both the outcomes is equal to 1.