From a pack of well-shuffled cards, two cards are drawn at random. Fin

1.	From a pack of well-shuffled cards, two cards are drawn at random. Find the probability that both the cards are diamonds when (i) the first card drawn is kept aside. (ii) the first card drawn is replaced in the pack.
Answer» In a pack of 52 cards, there are 13 diamond cards. Let event A: The first card drawn is a diamond card. ∴ P(A) = \(\frac {^{13}C_1}{^{52}C_1} = \frac {13} {52} = \frac 1 4\) (i) Let event B: The second card drawn is a diamond card. Since the first diamond card is kept aside, we now have 51 cards, out of which 12 are diamond cards. Probability that the second card is a diamond card under the condition that the first diamond card is kept aside in the pack =P(B/A) = \(\frac {^{12}C_1}{^{51}C_1} = \frac {4} {17}\) ∴ Required probability = P(A ∩ B) = P(B/A) . P(A) = 1/4 x 4/17 = 1/17 (ii) Let event B: The second card drawn is a diamond card. Since the first diamond card is replaced in the pack, we now again have 52 cards, out of which 13 are diamond cards. ∴ Probability that the second card is a diamond card under the condition that the first diamond card is replaced in the pack = P(B/A) = \(\frac {^{13}C_1}{^{52}C_1} = \frac {13} {52} = \frac 14\) Required probability = P(A ∩ B) = P(B/A) . P(A) = 1/4 x 1/4 = 1/16

From a pack of well-shuffled cards, two cards are drawn at random. Find the probability that both the cards are diamonds when (i) the first card drawn is kept aside. (ii) the first card drawn is replaced in the pack.

Answer»

In a pack of 52 cards, there are 13 diamond cards.

Let event A: The first card drawn is a diamond card.

∴ P(A) = \(\frac {^{13}C_1}{^{52}C_1} = \frac {13} {52} = \frac 1 4\)

(i) Let event B: The second card drawn is a diamond card.

Since the first diamond card is kept aside, we now have 51 cards, out of which 12 are diamond cards.

Probability that the second card is a diamond card under the condition that the first diamond card is kept aside in the pack

=P(B/A) = \(\frac {^{12}C_1}{^{51}C_1} = \frac {4} {17}\)

∴ Required probability = P(A ∩ B)

= P(B/A) . P(A)

= 1/4 x 4/17

= 1/17

(ii) Let event B: The second card drawn is a diamond card.

Since the first diamond card is replaced in the pack, we now again have 52 cards, out of which 13 are diamond cards.

∴ Probability that the second card is a diamond card under the condition that the first diamond card is replaced in the pack =

P(B/A) = \(\frac {^{13}C_1}{^{52}C_1} = \frac {13} {52} = \frac 14\)

Required probability = P(A ∩ B)

= P(B/A) . P(A)

= 1/4 x 1/4

= 1/16