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