Combined Probability of Unrelated Events

Date: 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/