Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
1229 Upvotes
Verified Answer
The correct answer is:
$\frac{2}{13}$
' $\mathrm{X}$ ' 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
$$
\begin{aligned}
& =\frac{{ }^4 \mathrm{C}_2 \times{ }^{48} \mathrm{C}_1}{{ }^{52} \mathrm{C}_2}=\frac{4 !}{2 ! 2 !} \times \frac{2 ! 50 !}{52 !}=\frac{1}{221} \\
& \mathrm{E}(\mathrm{X})=\Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}} \\
& =(0)\left(\frac{188}{221}\right)+(1)\left(\frac{32}{221}\right)+(2)\left(\frac{1}{221}\right)=\frac{2}{13}
\end{aligned}
$$
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
$$
\begin{aligned}
& =\frac{{ }^4 \mathrm{C}_2 \times{ }^{48} \mathrm{C}_1}{{ }^{52} \mathrm{C}_2}=\frac{4 !}{2 ! 2 !} \times \frac{2 ! 50 !}{52 !}=\frac{1}{221} \\
& \mathrm{E}(\mathrm{X})=\Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}} \\
& =(0)\left(\frac{188}{221}\right)+(1)\left(\frac{32}{221}\right)+(2)\left(\frac{1}{221}\right)=\frac{2}{13}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.