Two cards are drawn (without replacement) from a well shuffled deck of 52 cards. Find probability distribution and mean of number of cards numbered 4.

Hint: To find the probability distribution consider all the cases of 4 being drawn once twice.
Total no. of cards in the pack = 52
The number of cards numbered 4 in the pack of cards = 4
Let X be the discrete random variable denoting the number of cards numbered 4 when two cards are drawn without replacement.
X can take values 0, 1, 2.
Now,
Case 1: Probability of number 4 being drawn
$P\left( {X = 0} \right) = \dfrac{{{}^{48}{C_2}}}{{{}^{52}{C_2}}} = \dfrac{{188}}{{221}}$
Case 2: Probability of one 4 being drawn
$P\left( {X = 1} \right) = \dfrac{{{}^{48}{C_1} \times {}^4{C_1}}}{{{}^{52}{C_2}}} = \dfrac{{32}}{{221}}$
Case 3: Probability of two 4 being drawn
$P\left( {X = 2} \right) = \dfrac{{{}^4{C_2}}}{{{}^{52}{C_2}}} = \dfrac{1}{{221}}$
Therefore, the probability distribution is –

Now, mean of probability distribution –
$Mean = \sum {{x_i}{P_i}} $
$Mean = 0 \times \dfrac{{188}}{{221}} + 1 \times \dfrac{{32}}{{221}} + 2 \times \dfrac{1}{{221}}$
$Mean = \dfrac{{0 + 32 + 2}}{{221}} = \dfrac{{34}}{{221}}$

Note: To find the mean of probability distribution we first found the probability distribution across the values 0 to 2 and then applied summation to find the mean.