How do you solve this...?

Question

If you have to choose 5 cards out of a standard deck of cards (52 cards), what is the probability that you will get all face cards? What about, no face cards? Lastly, at least 1 face card?

please show steps

tell the number of ways for each

Answer 1

A standard 52 card deck has 12 face cards and 40 faceless cards. The total number of ways to draw 5 cards out of 52 is C(52, 5), where C(n, k) = n! / (k! (n - k)!). The total number of ways to do that ending up with exactly m face cards is...

C(12, m) * C(40, 5 - m)

... for m = 0, 1, 2, 3, 4, 5.

Answer 2

Well, for each suit there are 4 face cards...so 12 face cards in the entire deck.

Probability of getting all face cards:
12/52 * 11/51 * 10/50 * 9/49 * 8/48 = 0.0003047 (approx)

Probability of getting no face cards:
40/52 * 39/51 * 38/50 * 37/49 * 36/48 = 0.25318 (approx)

At least 1 face card - this depends when this face card is drawn...there's at least a 12/52 or 3/13 chance though.

Answer 3

first, if you select 5 cards out of 52 cards, the total combination is 52! / ((52-5)! * 5!) = 2,598,960.
There are 12 faces cards, the combination of geting 5 out of 12 is 12! / ((12-5)! * 5) = 792.
so the probability is 792 / 2,598,960.

again 2,598,960 possible combinations. there are 40 non-face cards. The combinations of geting 5 out of 40 is:
40!/((40-5)! * 5) = 658008.
so the probabiltiy is 658008 / 2,598,960.

At least one face card. This is a huge step.
combinations of 1 face cards and 4 non-face cards.
12 * (40*39*38*37)/4!= 91390

combinations of 2 faces card and 3 non-face cards
(12*11)/2 * (40*39*38)/3! = 652,080

combinations of 3 face cards and 2 non-face cards.
(12*11*10)/3! * (40*39)/2! = 171600

combinations of 4 face cards and 1 non-face cards.
(12*11*10*9)/4! * 40 = 198000

and combinations of 5 face card. I just found it the first step, which is 792

so the total is 91390 + 652,080 + 171600 + 198000 + 792
= 1113862.
So the probability is 1113862 / 2,598,960.

Answer 4

There are 12 face cards (K, Q, J).

P(all face) = 12C5 / 52C5

P(no face) = 40C5 / 52C5

P(at least 1 face) = 1 - P(no face)
= 1 - 40C5 / 52C5

Answer 5

23% Percent chance that you will get all face cards