Probability: Choosing From A Deck Of Cards

[Belfina]

Hey there! So the problem is like this:
We choose 10 random cards from a normal deck of cards(52 cards). What is the probability that we get:
a. 0 aces
b. maximum 3 aces
c. at least 1 ace and at least one face card

I'm unsure which formula I should use. I have thought that maybe the sample space is: 52!/10!(52-10)!..
But I still can't solve it

 

[HallsofIvy]

First, what you give is not the "sample space". A sample space is not a number!
Second, rather than using a "formula" you should be able to reason it out.

For the first one, of the 52 cards in a deck, 4 are aces, 48 are not. When you draw the first card, the probability it is not an ace is 48/52. If it is not an ace there are 51 cards left, 4 aces, 47 not. The probability the second card is not an ace is 47/51. Arguing in the same way, the probabilities that the third, fourth, fifth, sixth, seventh, eighth, ninth, and tenth cards are not aces are 46/50, 45/49, 44/48, 43/47, 42/46, 41/45, 40/44, and 39/43. The probability that, of all 10 cards, none is an ace is the product of those fractions, (48/52)(47/51)(46/50)(45/49)(44/48)(43/47)(42/46)(41/45)(40/44)(39/43).

b) A "maximum of three aces" means "0 aces or 1 ace or 2 aces or 3 aces". The probability of 0 aces is the calculation above. For one ace we can start with the assumption that the ace is the first card drawn. The probability the first card is an ace is 4/52. Then the probability the second is not an ace is 48/51, then 47/50, etc. The probability the first card drawn is an ace and the next nine are not is (4/52)(48/51)(47/50)(46/49)(45/48)(44/47)(43/46)(42/45)(41/44)(40/43).

Doing the same kind of calculation for the case that "the second card is an ace, all others not aces", is exactly the same. And, in fact, the probability the ace is in any given position, the other not aces, is exactly the same- the numerators and denominators are exactly the same, the numerators just in a different order. Since there are 10 places the ace could be multiply that by 10: 10(4/52)(48/51)(47/50)(46/49)(45/48)(44/47)(43/46)(42/45)(41/44)(40/43).

Similarly the probability the first two cards are aces and the other 8 not is (4/52)(3/51)(48/50)(47/49)(46/48)(45/47)(44/46)(43/45)(42/44)(41/43). But there are 10!/(2!)(8!)= 45 ways those two aces can appear in 10 cards so the probability of 2 aces and 8 non-aces in any order is 45(4/52)(3/51)(48/50)(47/49)(46/48)(45/47)(44/46)(43/45)(42/44)(41/43)

The probability the first three cards are aces and the other 7 are not is (4/52)(3/51)(2/50)(48/49)(47/48)(46/47)(45/46)(44/45)(43/44)(43/44)(42/43). But now there are 10!/3!7!= 120. The probability of three aces and 7 non-aces in any order is 120(4/52)(3/51)(2/50)(48/49)(47/48)(46/47)(45/46)(44/45)(43/44)(43/44)(42/43).

The probability of "a maximum of 3 aces" is the sum of those.

"At least one ace and at least one face card" is best done by calculating the probability that does NOT happen, then subtracting from 1. So first calculate the probability of "no ace", then calculate the probability or "no face card" and add. But that would include "no ace" and "no face card" twice so you need to subtract of the probability or "no ace and no face card" before subtracting from 1.