# Thread: [SOLVED] Depend event 2

1. ## [SOLVED] Depend event 2

this is another problem for me.. i always stuck on playing card problems:
hope some one can guide me on this

Question:
From a pack of 52 playing card, two card are selected without replacement. What is the probability that:
(a) both card are diamond (ans: 1/17)
(b) both card are same suit(ans: 4/17)
(c) one card is a spade and the other a club (ans: 13/102)
(d) both card are different suits? (ans: 13/17)

2. Hello,nikk!

From a pack of 52 playing card, two card are selected without replacement.
What is the probability that:

(a) both cards are Diamonds? (ans: 1/17)

The first card must be a Diamond: .probability = $\tfrac{13}{52}$

. . There are 13 Diamonds among the 52 cards.

The second card must be a Diamond: .probability = $\tfrac{12}{51}$

. . There are 12 remaining Diamonds among the remaining 51 cards.

Therefore: . $P(\text{both Diamonds}) \:=:\frac{13}{52}\cdot\frac{12}{51} \:=\:\frac{1}{17}$

(b) both cards are same suit ? (ans: 4/17)

The first card can be ANY card: . $\tfrac{52}{52} \,=\,1$

The second card must match that suit: . $\tfrac{12}{51}$

. . There are 12 remaining cards of that suit.

Therefore: . $P(\text{same suit}) \:=\:1\cdot\frac{12}{51} \:=\:\frac{4}{17}$

(c) one card is a Spade and the other a Club? (ans: 13/102)
There are two cases to consider . . .

(1) The first is a Spade: . $\tfrac{13}{52}$
. . .The second is a Club: . $\tfrac{13}{51}$

(2) The first is a Club: . $\tfrac{13}{52}$
. . .The second is a Spade: . $\tfrac{13}{51}$

Therefore: . $P(\text{Spade \& Club}) \:=\:\frac{13}{52}\cdot\frac{13}{51} + \frac{13}{52}\cdot\frac{13}{51} \;=\;\frac{13}{102}$

(d) both cards are different suits? (ans: 13/17)

The first card can be ANY card: . $\tfrac{52}{52} \,=\,1$

The second card must not match the first: . $\frac{39}{51}$

. . There are 39 cards of another suit among the 51 remaining cards.

Therefore: . $P(\text{different suits}) \:=\:1\cdot\frac{39}{51} \:=\:\frac{13}{17}$