Combinatronics, drawing exactly 1 ace from a card deck

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1. Sep 8, 2015

whitejac

1. The problem statement, all variables and given/known data
In a 52 card deck, if you draw 5 cards find the probability of drawing exactly one ace.

2. Relevant equations
(n k) = n!/k!(n-k)!
P(A) = |A|/|S|

3. The attempt at a solution
So I took the logic of a different example we had that stated it like this - If we have 5 options but only 1 of them can be an ace then we have (5 1). This gives a very very unrealistic problem as the sample space is (52 5) yielding me 5/2598960 to equal roughly 1.62*10^-6

So I took a different logic stating that if we have 4 cards that are aces then 48 of them are not. Then we may choose 4 of these cards for (48 4).

My question is which one of these or are neither of them a very profficient way to approach these? I've seen it put where you can logically construct these situations using something to the effect of:
(n1 k1)(n2 k2) and this would produce the appropriate number for your |A| because you construct it through multiplecation. ie (48 1) = (12 1)(4 1).

2. Sep 8, 2015

SteamKing

Staff Emeritus
"Combinatronics" is randomly putting together electronic components.

"Combinatorics" is that branch of mathematics which studies the combinations of objects drawn from a finite set.

3. Sep 8, 2015

HallsofIvy

In a deck of 52 cards, 4 of them aces, so 4/52= 1/13 of them are aces. What is the probability that the first card drawn is an ace? If that happens then there are 51 cards left, 3 of them aces and 48 "non-aces". What are the probabilities that each of the next 4 cards drawn are NOT aces? The product of those is the probability of first drawing an ace and then four non-aces.

What is the probability that the first card drawn is NOT an ace, the second is an ace, and the next three are not aces? The product of those is the probability that the second card drawn is an ace and the four others are non-aces. The point is that you should see that while the individual fractions involved are different, the 5 numerators and 5 denominators are the same as before, just in different positions, so the product is the same.

That is, the probability of drawing a single ace and four non-aces, in any specific order, is the same whatever the order. And there are 5 such orders:
ANNNN, NANNN, NNANN, NNNAN, and NNNNA, where "A" represents an ace, "N" a non-ace.

4. Sep 8, 2015

whitejac

That was a very concise and intuitive explanation, thank you!

I'm a bit curious if there were not a better way to do it using combinations. I can see the factorials building up, but I'll need to spend more time playing with them before I think I'll really grasp it. I liked your approach though as it does more to derive the equations than simply memorize a tactic.

5. Sep 8, 2015

Ray Vickson

Sure: use the "hypergeometric distribution".

In this case you have 52 items, of which 4 are of type I (Aces) and 48 are of type II (non-Aces). You draw n = 5 items without replacement, and want the probability that your sample contains k = 1 item of type I. Google 'hypergeometric distribution' to see the relevant formulas.