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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.

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} \\

$