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Question 1087657: My Question: If THREE cards are dealt from a 52 card deck, what is the probability that at least one of those cards will be an Ace, Jack, or Seven?
My Thoughts: There are 4 Aces, 4 Jacks, and 4 Sevens in a 52 card deck. If you are dealt one card, you have a 12/52 or 23.1% chance that the card dealt is an Ace, Jack, or Seven. But I have no idea what the next step is when figuring out the probability if 3 cards are dealt. If I did 12/52+12/51+12/50, then the answer would be 70.6% which seems too high. PLEASE HELP!
Found 2 solutions by natolino_2017, math_helper:
Answer by natolino_2017(77)   (Show Source):
You can put this solution on YOUR website! In this case we can use Hypergeometric Distribution with N=52 (total number of cards), n= 3 (number of dealt cards), d = 4 + 4 + 4 = 12 (total working cards)
P(X=x) = (dCx)* ((N-d)C(n-x))/(NCn).
P(X>=1) = 1 - P(x=0) (using the complement rule).
Using the general term with x=0, P(x=0) = (12C0)(40C3)/(52C3) = 38/85.
So the answer is 1 - 38/85 = 47/85 =0,553
@natolino_
Answer by math_helper(2461)  (Show Source):
You can put this solution on YOUR website! A = { A, J, or 7 drawn }
B = { all the other cards }
So yes, there are 12 cards of interest. That means there are 40 other cards.
Pr(A) = 1 - Pr(only cards from B are drawn)
= 1 - (40/52)*(39/51)*(38/50)
= 1 - (38/(17*5))
~= 
( ~= used to indicate approximation)
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