Question

Two cards are drawn from a fifty-two-card deck. What is the probability that the draw will yield an ace and a face card?

Answer 1

I havent worked with probability in like a year XD i forgot how to :P

Answer 2

Think of it like this. There are 4 face/ace cards to every suit.

Answer 3

That is true

Answer 4

4 cards times 4 gives us 16. There are 16 out of 56 cards that are face cards.

Answer 5

I mean 52.

Answer 6

yeah I know

Answer 7

there are a couple ways to do this
but first we need to know does it mean "and ace and then a face card" which is easy
or does it mean "one of the two cards is and ace, the other is a face card" which is somewhat harder

Answer 8

Sorry ^^''.

Answer 9

ok

Answer 10

And true, @satellite73.

Answer 11

But if he meant the way I did it, then its a simple 16:52

Answer 12

oh no

Answer 13

But its not done yet. Find the common factor.

Answer 14

Did I mess up somewhere? o.o''

Answer 15

first card is an ace
\[\frac{4}{52}\]
second card is a face card given that the first card is an ace
\[\frac{12}{51}\]
AND
\[\frac{4}{52}\times \frac{12}{51}\]

Answer 16

that is "first ace, then face card" which might not be what the question is asking

Answer 17

one is a face card, the other is an ace
\[\frac{4\times 12}{\binom{52}{2}}\]

Answer 18

There is only an ace and a face card. I just asked my teacher

Answer 19

Only one ace and one face?

Answer 20

yes

Answer 21

\(\binom{52}{2}=\frac{52\times 51}{2}=1326\) your answer is
\[\frac{4\times 12}{1326}\]

Answer 22

Alright Satellite how did you get that answer

Answer 23

the denominator is all the ways you can choose two cards out of fifty two
i.e. "fifty two choose 2" written as \(\binom{52}{2}\) or sometimes \(_{52}C_2\) or even \(^{52}C_2\) and computed via
\[\frac{52\times 51}{2}\]

Answer 24

the numerator is the number of ways you can pick one ace out of four, which is \(4\) times the number of ways you can pick 1 facecard out of 12, which is 12

Answer 25

Like @satellite73 said,
\[P(1\text{ ace}\cap1\text{ face})=\frac{P(1\text{ ace})\cdot P(1\text{ face})}{P(2\text{ cards from deck}}=\frac{\binom 41\binom{12}1}{\binom{52}2}\]

Answer 26

thanks