Probabilty question

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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 if a queen?

The answer is 11/663, but I'm not sure what the steps were

I've tried ((12/52)(4/41))+((12/52)(3/41)) and end up with (7/221), and I've also tried (4/52)(11/51) which gives me the right answer, but the problem is that the sequence of picking cards is wrong. I've no idea where I went wrong.

Please help!
 
Let Q<SUB>1</SUB> be the event that the first card drawn is a Queen.
Let (Q<SUB>1</SUB>)<SUP>c</SUP> be the event that the first card drawn is a face card not a queen.
Let F<SUB>1</SUB> be the event that the first card drawn is a face card. Thus
\(\displaystyle \begin{array}{l}
P(Q_2 \cap F_1 ) = P\left( {Q_2 \cap \left[ {Q_1 \cup Q_1^c } \right]} \right) \\
= P(Q_2 \cap Q_1 ) + P(Q_2 \cap Q_1 ^c ) \\
= P(Q_2 |Q_1 )P(Q_1 ) + P(Q_2 |Q_1 ^c )P(Q_1 ^c ) \\
= \frac{3}{{51}}\frac{4}{{52}} + \frac{4}{{53}}\frac{8}{{52}} = \frac{{11}}{{663}} \\
\end{array}\).
 
Hello, Amphitrite!

Here is pka's explanation . . . in baby-talk.

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?

The answer is 11/663
Click to expand...
There are two cases to consider:
\(\displaystyle \;\;\)(1) The first card is a Queen.
\(\displaystyle \;\;\)(2) The first card is not a Queen.

Case 1: The first card drawn is a Queen.
. . We have: \(\displaystyle P(\)first is Q\(\displaystyle )\:=\:\frac{4}{52}\)
. . Then: \(\displaystyle \,P(\)second is Q\(\displaystyle )\:=\:\frac{3}{51}\)
Hence: \(\displaystyle P(\)Q, then Q\(\displaystyle )\:=\:\left(\frac{4}{52}\right)\left(\frac{3}{51}\right) \:=\:\frac{3}{663}\)

Case 2: The first card is a face card, not a Queen.
. . We have: \(\displaystyle \,P(\)first is face, not Q\(\displaystyle )\:=\:\frac{8}{52}\)
. . Then: \(\displaystyle \,P(\)second is Q\(\displaystyle )\:=\:\frac{4}{51}\)
Hence: \(\displaystyle P(\) face not Q, then Q\(\displaystyle )\:=\;\left(\frac{8}{52}\right)\left(\frac{4}{51}\right)\:=\;\frac{8}{663}\)

Therefore: \(\displaystyle \.P(\)face, then Q\(\displaystyle )\:=\:\frac{3}{663}\,+\,\frac{8}{663}\:=\:\frac{11}{663}\)
 
hello what if the question was changed to

a.) What is the probability that the first card is a face card OR the second card if a queen?
b.) What is the probability that the first card is a face card and the second card NOT a queen?
 
There is more than one way to answer this.

The more intuitive one is to calculate three probabilities.

First card face and second card not queen

First card face and second card queen

First card not face and second card queen

The more computaionally economical way is to subtract the probability of first card not face and second card not queen from 1.

Try both ways to see you get the same answer.
 
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