Question

Three cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the mean and variance of number of red cards. 

Answer 1

It is given that the cards are drawn successively with replacement so the the events are independent. Therefore, the drawing of the cards follow binomial distribution.

​Probability of drawing a red card = p = \[\frac{26}{52} = \frac{1}{2}\]

∴ q = 1 − p = \[1 - \frac{1}{2} = \frac{1}{2}\]

Also, n = 3
Let X be the random variable denoting the number of red cards drawn from a well shuffled pack of 52 cards.

\[\therefore P\left( X = r \right) =^ {3}{}{C}_r \left( \frac{1}{2} \right)^{3 - r} \left( \frac{1}{2} \right)^r =^{3}{}{C}_r \left( \frac{1}{2} \right)^3 , r = 0, 1, 2, 3\] 

 Probability of drawing no red ball =\[P\left( X = 0 \right) = ^{3}{}{C}_0 \left( \frac{1}{2} \right)^3 = \frac{1}{8}\]

Probability of drawing one red ball =  \[P\left( X = 1 \right) = ^{3}{}{C}_1 \left( \frac{1}{2} \right)^3 = \frac{3}{8}\]

Probability of drawing two red balls = \[P\left( X = 2 \right) = ^{3}{}{C}_2 \left( \frac{1}{2} \right)^3 = \frac{3}{8}\]

Probability of drawing three red balls = \[P\left( X = 3 \right) = ^{3}{}{C}_3 \left( \frac{1}{2} \right)^3 = \frac{1}{8}\]

Thus, the probability distribution of X is as follows:

``

xi	pi	pixi	`p_ix_i^2`
0	`1/8`	0	0
1	`3/8`	`3/8`	`3/8`
2	`3/8`	`6/8`	`12/8`
3	`1/8`	`3/8`	`9/8`
		`sum p_ix_i = 12/8`	`sum p_ix_i^2 = 3`

Mean of X 

\[= \sum^{}_{} p_i x_i = \frac{12}{8} = \frac{3}{2}\]

Variance of X 

\[= \sum^{}_{} p_i x_i^2 - \left( \text{ Mean}  \right)^2 = 3 - \frac{9}{4} = \frac{3}{4}\]