Question
Two cards are drawn simultaneously (without replacement) from a well-shuffled pack of 52 cards. Find the mean and variance of the number of red cards.
Let X be a random variable which can take values
0, 1, 2 where X is the no. of red cards selected
$\therefore$ X = 0 means 0 red card
P(X = 0) =
$\frac{26_{\text{c}_{2}}}{\text{52}_{\text{c}_{2}}}=\frac{26\times25}{52\times51}=\frac{25}{102}$P(X = 1) =
$\frac{26_{\text{c}_{1}}\times26_{\text{c}_{1}}}{\text{52}_{\text{c}_{2}}}=\frac{26\times26\times2}{52\times51}=\frac{52}{102}$P(X = 2) =
$\frac{26_{\text{c}_{2}}}{\text{52}_{\text{c}_{2}}}=\frac{26\times25}{52\times51}=\frac{52}{102}$Probability distribution of random variable X is
| X | 0 | 1 | 2 |
| P(X) | 25/102 | 52/102 | 25/102 |
Mean = $\Sigma$ x P(X) =
$\frac{52+50}{102}=1$variance =
$\Sigma\text{x}^{2}\text{P(X)-(}\Sigma\text{xP(X))}^{2}=\frac{152}{102}-1=\frac{50}{102}\text{ OR }\frac{25}{51}$.Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$
Verify that (adjoint A)A = |A|I = A (adjoint A) for the above matrices.