Four cards are drawn at random from a pack of playing cards. The probability of getting exactly two cards…

We have $52$ pack of cards.

$$\begin{gathered} 52\text{ cards}\to \left\{\begin{array}{l}\text{Club}\to 13 \\ \text{Diamond}\to 13\\ \text{Heart}\to 13 \\ \text{Spades}\to 13\end{array}\right. \end{gathered}$$

Now, for the favourable case we will select any suit and pick two cards from that, and then we will select another suit from the remaining $3$ suits and pick two cards.

Favourable ways$={}^{4}C_1\times {}^{13}C_2\times {}^{3}C_2\times {}^{13}C_1\times {}^{13}C_1$.

Total ways$={}^{52}C_4$.

Required probability$=\frac{{}^{4}C_1\times {}^{13}C_2\times {}^{3}C_2\times {}^{13}C_1\times {}^{13}C_1}{{}^{52}C_4}$

$=\frac{4\times \frac{13\times 12}{2}\times 3\times 13\times 13}{\frac{52\times 51\times 50\times 49}{4\times 3\times 2\times 1}}=\frac{6\times 13\times 12\times 13\times 13\times 24}{52\times 51\times 50\times 49}$

$=\frac{6\times 2\times 13\times 13\times 6}{17\times 25\times 49}=\frac{18\times 4\times 13\times 13}{17\times 25\times 49}=\frac{72\times 169}{425\times 49}$