Search any question & find its solution
Question:
Answered & Verified by Expert
Two cards are drawn from a pack of 52 playing cards one after the other. If $p_1$ is the probability of getting a queen in the first draw and a diamond card in the second draw when the first card drawn is replaced and $p_2$ is the probability of the same event when the first card drawn is not replaced. Then $\frac{\mathrm{p}_1}{\mathrm{p}_2}=$
Options:
Solution:
1265 Upvotes
Verified Answer
The correct answer is:
1
Case 1: with replacement
$\mathrm{P}_1=\mathrm{P}$ (First queen and second diamond $)$
$$
=\frac{4}{52} \times \frac{13}{52}=\frac{1}{52}
$$
Case 2 : without replacement
$\mathrm{P}_2=\mathrm{P}$ (First queen and second diamond)
$={ }^2 \mathrm{P}($ First diamond queen and second diamond $)$
$+\mathrm{P}$ (First other than queen and second diamond)
$$
=\frac{1}{52} \times \frac{12}{51}+\frac{3}{52} \times \frac{13}{51}=\frac{51}{52 \times 51}=\frac{1}{52}
$$
$$
\therefore \frac{\mathrm{P}_1}{\mathrm{P}_2}=1
$$
$\mathrm{P}_1=\mathrm{P}$ (First queen and second diamond $)$
$$
=\frac{4}{52} \times \frac{13}{52}=\frac{1}{52}
$$
Case 2 : without replacement
$\mathrm{P}_2=\mathrm{P}$ (First queen and second diamond)
$={ }^2 \mathrm{P}($ First diamond queen and second diamond $)$
$+\mathrm{P}$ (First other than queen and second diamond)
$$
=\frac{1}{52} \times \frac{12}{51}+\frac{3}{52} \times \frac{13}{51}=\frac{51}{52 \times 51}=\frac{1}{52}
$$
$$
\therefore \frac{\mathrm{P}_1}{\mathrm{P}_2}=1
$$
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.