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?

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- Jun 22nd 2006, 10:55 PMstuck_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, 01:49 AMCaptainBlackQuote:

Originally Posted by**stuck_fugu**

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

$

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

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

$\displaystyle

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

$

$\displaystyle

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

$

$\displaystyle

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

$

$\displaystyle

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

$.

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

RonL - Jun 23rd 2006, 04:24 AMSoroban
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.

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

**[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).

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

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