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

Hint: For solving this problem, we start by calculating the probabilities of drawing 0 red cards, 1 red card and 2 red cards using the formula for number of ways of selecting r objects from n objects,${}^{n}{{C}_{r}}$. Then using these values, we calculate the mean and variance using the formulas $Mean=\sum\limits_{x}{xP\left( X=x \right)}$ and $Variance=\sum\limits_{x}{{{x}^{2}}P\left( X=x \right)}-{{\left( \sum\limits_{x}{xP\left( X=x \right)} \right)}^{2}}$.

Complete step by step answer:
First, to solve the problem, we need to know the number of black and red cards present in a pack of 52 cards.
Number of red cards in a well-shuffled pack of 52 cards = 26.
Number of black cards in a well-shuffled pack of 52 cards = 26.
We are given that two cards are drawn simultaneously without replacement. We need to find the mean and variance of the number of red cards.
When two cards are drawn from the pack, we have three possibilities.
1. Drawing 1 red card and drawing 1 black card.
2. Drawing 2 red cards.
3. Drawing 0 red cards, that is drawing 2 black cards
Let us consider the event of drawing a number of red cards as X.
Then the possible values of X are 0, 1, 2.
Now let us calculate the probabilities when X=0, X=1 and X=2.
When X=0,
$P\left( X=0 \right)=\dfrac{\text{Number of ways of drawing 2 black cards}}{\text{Number of ways of drawing 2 cards}}$
Let us consider the formula for number of ways of selecting r objects from n objects =${}^{n}{{C}_{r}}$
$\begin{align}
  & \Rightarrow P\left( X=0 \right)=\dfrac{{}^{26}{{C}_{2}}}{{}^{52}{{C}_{2}}} \\
 & \Rightarrow P\left( X=0 \right)=\dfrac{\dfrac{26!}{24!\times 2!}}{\dfrac{52!}{50!\times 2!}} \\
 & \Rightarrow P\left( X=0 \right)=\dfrac{25\times 26}{51\times 52} \\
 & \Rightarrow P\left( X=0 \right)=\dfrac{25}{51\times 2} \\
 & \Rightarrow P\left( X=0 \right)=\dfrac{25}{102} \\
\end{align}$
When X=1,
$P\left( X=1 \right)=\dfrac{\text{Number of ways of drawing 1 red and 1 black cards}}{\text{Number of ways of drawing 2 cards}}$
So, applying the formula for selecting 1 red and 1 black cards, we get
$\begin{align}
  & \Rightarrow P\left( X=1 \right)=\dfrac{{}^{26}{{C}_{1}}\times {}^{26}{{C}_{1}}}{{}^{52}{{C}_{2}}} \\
 & \Rightarrow P\left( X=1 \right)=\dfrac{\dfrac{26!}{25!\times 1!}\times \dfrac{26!}{25!\times 1!}}{\dfrac{52!}{50!\times 2!}} \\
 & \Rightarrow P\left( X=1 \right)=\dfrac{26\times 26\times 2}{51\times 52} \\
 & \Rightarrow P\left( X=1 \right)=\dfrac{26}{51} \\
\end{align}$
Similarly, when X=2
$P\left( X=2 \right)=\dfrac{\text{Number of ways of drawing 2 red cards}}{\text{Number of ways of drawing 2 cards}}$
So, applying the formula for selecting 1 red and 1 black cards, we get
$\begin{align}
  & \Rightarrow P\left( X=2 \right)=\dfrac{{}^{26}{{C}_{2}}}{{}^{52}{{C}_{2}}} \\
 & \Rightarrow P\left( X=2 \right)=\dfrac{\dfrac{26!}{24!\times 2!}}{\dfrac{52!}{50!\times 2!}} \\
 & \Rightarrow P\left( X=2 \right)=\dfrac{25\times 26}{51\times 52} \\
 & \Rightarrow P\left( X=2 \right)=\dfrac{25}{51\times 2} \\
 & \Rightarrow P\left( X=2 \right)=\dfrac{25}{102} \\
\end{align}$
So, now we have the probabilities for X as


Let us consider the formula for calculating mean.
$Mean=\sum\limits_{x}{xP\left( X=x \right)}$
Here x takes values 0, 1 and 2.
So, applying the formula we get,
$\begin{align}
  & \Rightarrow Mean=\sum\limits_{x}{xP\left( X=x \right)}=0\left( \dfrac{25}{102} \right)+1\left( \dfrac{26}{51} \right)+2\left( \dfrac{25}{102} \right) \\
 & \Rightarrow Mean=\dfrac{26}{51}+\dfrac{25}{51} \\
 & \Rightarrow Mean=1...................\left( 1 \right) \\
\end{align}$
Now, let us consider the formula for calculating the variance.
$\begin{align}
  & Variance=\sum\limits_{x}{{{x}^{2}}P\left( X=x \right)}-{{\left( \sum\limits_{x}{xP\left( X=x \right)} \right)}^{2}} \\
 & Variance=\sum\limits_{x}{{{x}^{2}}P\left( X=x \right)}-{{\left( Mean \right)}^{2}} \\
\end{align}$
Now let us apply this formula and calculate the variance.
$\begin{align}
  & \Rightarrow Variance=\sum\limits_{x}{{{x}^{2}}P\left( X=x \right)}-{{\left( Mean \right)}^{2}} \\
 & \Rightarrow Variance=\left( {{\left( 0 \right)}^{2}}\left( \dfrac{25}{102} \right)+{{\left( 1 \right)}^{2}}\left( \dfrac{26}{51} \right)+{{\left( 2 \right)}^{2}}\left( \dfrac{25}{102} \right) \right)-{{\left( 1 \right)}^{2}} \\
 & \Rightarrow Variance=\left( 0+\left( 1 \right)\left( \dfrac{26}{51} \right)+\left( 4 \right)\left( \dfrac{25}{102} \right) \right)-1 \\
 & \Rightarrow Variance=\left( \dfrac{26}{51}+\dfrac{100}{102} \right)-1 \\
 & \Rightarrow Variance=\left( \dfrac{26}{51}+\dfrac{50}{51} \right)-1 \\
 & \Rightarrow Variance=\dfrac{76}{51}-1 \\
 & \Rightarrow Variance=\dfrac{76-51}{51} \\
 & \Rightarrow Variance=\dfrac{25}{51} \\
 & \Rightarrow Variance=0.49...................\left( 2 \right) \\
\end{align}$
Using equations (1) and (2) we get Mean = 1 and Variance=0.49.
Hence the answer is Mean = 1 and Variance=0.49.

Note:
The place where most of the students make mistake while solving this question is taking the wrong formula for number of ways of selecting r objects from n objects as ${}^{n}{{P}_{r}}$ instead of taking ${}^{n}{{C}_{r}}$. So, one needs to remember that the formula for number of ways of selecting r objects from n objects is ${}^{n}{{C}_{r}}$, while ${}^{n}{{P}_{r}}$ is for selecting and arranging r objects from n objects.