Two cards are drawn simultaneously (without replacement) from a well-shuffled deck of 52 cards. Find the mean and variance of number of red cards.

Firstly, find the probability distribution table of number of red cards. Then, using this table, find the mean and variance.
Let X be the number of red cards. Then, X can take values 0, 1 and 2.

$$\begin{gathered} P(X=0)=P(having no red card) \\ =\frac{{}^{26}\mathrm{C}_2}{{}^{52}{\mathrm{C}}_2}=\frac{26\times 25/(2\times 1)}{52\times 51/(2\times 1)}=\frac{1}{2}\times \frac{25}{51} \\ =\frac{25}{102} \\ \mathrm{P} (\mathrm{X}=1)=\mathrm{P} (\mathrm{having} \mathrm{one} \mathrm{red} \mathrm{card}) \\ =\frac{{}^{26}{\mathrm{C}}_1\times {}^{26}{\mathrm{C}}_1}{{}^{52}{\mathrm{C}}_2} \\ =\frac{26\times 26\times 2}{52\times 51}=\frac{26}{51} \\ \mathrm{P} (\mathrm{X}=2)=\mathrm{P} (\mathrm{having} \mathrm{two} \mathrm{red} \mathrm{cards}) \\ =\frac{{}^{26}{\mathrm{C}}_2}{{}^{52}{\mathrm{C}}_2}=\frac{26\times 25/2\times 1}{52\times 51/2\times 1}=\frac{25}{102} \end{gathered}$$

$\therefore$ The probability distribution of number of red cards is given below

Now, we know that, mean $=\sum \mathrm{x}. \mathrm{P}\left(\mathrm{X}\right)$

And variance $=\sum {\mathrm{x}}^2.\mathrm{P}\left(\mathrm{x}\right)-{\left[\sum \mathrm{x}.\mathrm{P}\left(\mathrm{X}\right)\right]}^2$

Now, P(X = 0) = p (having no red card)

$$\begin{gathered} \mathrm{Mean}=\sum \mathrm{X}.\mathrm{P}\left(\mathrm{X}\right) \\ =0+\frac{26}{51}+\frac{25}{51}+\frac{51}{51}=1 \\ \mathrm{Variance}=\sum {\mathrm{X}}^2.\mathrm{P}\left(\mathrm{X}\right)-{[\sum \mathrm{X}.\mathrm{P}(\mathrm{X}\left)\right]}^2 =\left[0+\frac{26}{51}+\frac{50}{51}\right]-{\left(1\right)}^2 =\frac{76}{51}-1=\frac{76-51}{51}=\frac{25}{51} \end{gathered}$$