|By Mooses41 (Mooses41) on Tuesday, June 08, 2004 - 01:08 am: Edit|
hey everyone,im having trouble with this one math problem, could someone please show me how to do it?
13 cards are delt from a standard deck. whats the probability of getting at least 3 aces? leave your answer in terms of combinations
thanks so much for the help
|By Homer2568 (Homer2568) on Tuesday, June 08, 2004 - 10:53 am: Edit|
1-(4C2)(48C11)+(4C1)(48C12)+(48C13) all divided by (52C13)....this is the complement rule...so instead of figuring out the probability of getting at least 3 aces...at least 4...5...6...so on you do 1 minus the probability of getting 2 or 1 or 0 aces that way it makes it shorter and it workes nicer.
|By Vanessa1378 (Vanessa1378) on Tuesday, June 08, 2004 - 01:04 pm: Edit|
Wouldn't it just be (4C4)(48C9)+(4C3)(48C10) since there are only 4 aces.
|By Xiggi (Xiggi) on Tuesday, June 08, 2004 - 01:56 pm: Edit|
I would answer
P(exactly three aces) = C(4,3) * C(48,10) / C(52,13) = 0.04120048 or 0.04
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