- #1
whitejac
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Homework Statement
In a 52 card deck, if you draw 5 cards find the probability of drawing exactly one ace.
Homework Equations
(n k) = n!/k!(n-k)!
P(A) = |A|/|S|
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).