# [SOLVED] Urgent help required: Probability

• Jun 22nd 2006, 11:55 PM
stuck_fugu
[SOLVED] Urgent help required: Probability
Two cards are drawn without replacement from a standard deck of 52 cards. What is the probability that the first card is a face card and the second card is a queen?

I keep getting 7/221

but the answer says it's 11/663.

anyone know?
• Jun 23rd 2006, 02:49 AM
CaptainBlack
Quote:

Originally Posted by stuck_fugu
Two cards are drawn without replacement from a standard deck of 52 cards. What is the probability that the first card is a face card and the second card is a queen?

I keep getting 7/221

but the answer says it's 11/663.

anyone know?

$
P(1st\ face \land 2nd\ Q)=P(1st\ K \lor N)P(2nd\ Q|1st\ K \lor N)$
$+ P(1st\ Q)P(2nd\ Q|1st\ Q)
$

(where $\land$ should be read as "and", and $\lor$ should be read as "or".

Now as there are four of each value of card in a deck and 52 cards:

$
P(1st\ K \lor N)=\frac{8}{52}
$

$
P(1st\ Q)=\frac{4}{52}
$

$
P(2nd\ Q|1st\ K \lor N)=\frac{4}{51}
$

$
P(2nd\ Q|1st\ Q)=\frac{3}{51}
$
.

And I'm sure you can do the final sum for yourself.

RonL
• Jun 23rd 2006, 05:24 AM
Soroban
Hello, stuck_fugu!

Quote:

Two cards are drawn without replacement from a standard deck of 52 cards.
What is the probability that the first card is a face card and the second card is a queen?

There are two possible cases:

[1] The first card is a Queen and the second is a Queen.
. . . There are 4 choices for the first card; 3 choices for the second.

. . . $P(\text{Q, then Q})\:=\:\frac{4}{52}\cdot\frac{3}{51}\;=\;\frac{3} {663}$

[2] The first card is a Face card (not a Queen) and the second is a Queen.
. . . There are 8 choices for the first card (4 Jacks, 4 Kings)
. . . and 4 choices for the second card (4 Queens).

. . . $P(\text{J or K, then Q})\:=\:\frac{8}{52}\cdot\frac{4}{51}\:=\:\frac{8} {663}$

Therefore: . $P(\text{Face and Q})\:=\;\frac{3}{663} + \frac{8}{663}\:=\:\frac{11}{663}$