Let two cards are drawn at random from a pack of 52 playing cards. Let \(\mathrm{X}\) be the number of aces obtained. Then the values of \(\mathrm{E}(\mathrm{X})\) is

' \(\mathrm{X}\) ' can take values 0.1 .2 .
Probability of getting no ace card.
\(
=\frac{{ }^{48} \mathrm{C}_2}{{ }^{52} \mathrm{C}_2}=\frac{48 !}{2 ! 46 !} \times \frac{2 ! 50 !}{52 !}=\frac{188}{221}
\)
Probability of getting 1 ace card
\(
=\frac{{ }^4 \mathrm{C}_1 \times{ }^{48} \mathrm{C}_1}{{ }^{52} \mathrm{C}_2}=\frac{4 \times 48}{52 !} 2 ! 50 !=\frac{32}{221}
\)
Probability of getting 2 ace cards
\(
\begin{aligned}
& =\frac{{ }^4 \mathrm{C}_2 \times{ }^{48} \mathrm{C}_1}{{ }^{52} \mathrm{C}_2}=\frac{4 !}{2 ! 2 !} \times \frac{2 ! 50 !}{52 !}=\frac{1}{221} \\
& \mathrm{E}(\mathrm{X})=\Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}} \\
& =(0)\left(\frac{188}{221}\right)+(1)\left(\frac{32}{221}\right)+(2)\left(\frac{1}{221}\right)=\frac{2}{13}
\end{aligned}
\)