Combined Probability of Unrelated EventsDate: 7/23/96 at 0:25:42 From: Anonymous Subject: Combined Probability of Unrelated Events - 5 Cards Five cards are drawn without replacement from a standard deck of playing cards. Next, five cards are drawn without replacement from a second standard deck of playing cards. And finally, five cards are drawn without replacement from a third standard deck of playing cards. What is the probability that the Ace of Spades will appear once? What is the probability that the Ace of Spades will appear twice? What is the probability that the Ace of Spades will appear three times? Is this another Hypergeometric Distribution problem? I know NCx = N! / (x! * (N-x)!) = 52! / (5! * (52-5)!) = 2,598,960 But where do I go from here? This is somewhat confusing. Thanks Date: 7/23/96 at 16:10:20 From: Doctor Anthony Subject: Re: Combined Probability of Unrelated Events - 5 Cards The chance that the Ace of Spades is chosen when five cards are dawn from one pack is (1C1*51C4)/(52C5) = 0.09615 Chance that it is not selected is 1 - 0.09615 = 0.903846 So we have three trials, in each of which there is a probability of success of 0.09615, and a chance of failure of 0.903846 This is a binomial probability problem, and we have : Probability of one success = 3C1*0.09615*0.903846^2 = 0.235646 Probability of two successes = 3C2*0.09615^2*0.903846 = 0.0250677 Probability of three successes = 3C3*0.09615^3 = 0.0008889 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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