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June 7th, 2012, 11:31 AM  #1 
Member Joined: Dec 2010 From: Miami, FL Posts: 96 Thanks: 0  Probability of Drawing Card
I have a 40 card deck with five cards colored red. What is the probability that you'll draw at least one red card in a hand of six?

June 7th, 2012, 12:03 PM  #2 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361  Re: Probability of Drawing Card
Let x be the amount of cards drawn from the deck. Then p(x = 0) + p(x = 1) + p(x = 2) + p(x = 3) + p(x = 4) + p(x = 5) + ... = 1 All possible outcomes add to 1. p(x > 5) = 0; you cannot draw more then 5 reds since there are only five. Now, we are left with p(x = 0) + p(x = 1) + p(x = 2) + p(x = 3) + p(x = 4) + p(x = 5) = 1 i.e. p(x = 0) + [p(x = 1) + p(x = 2) + p(x = 3) + p(x = 4) + p(x = 5)] = 1 Also: p(x = 0) + p(x >= 1) = 1 yields p(x >= 1) = 1  p(x = 0) Can you find the RHS? 
June 7th, 2012, 01:08 PM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Re: Probability of Drawing Card Hello, guynamedluis! Quote:
The opposite of "at least one red card" is "no red cards."  
June 7th, 2012, 11:14 PM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Probability of Drawing Card soroban, it appears you have accidentally used permutations instead of combinations to get the number of hands described. Your end result is correct though since the difference between the two functions cancels out in the division. 
June 8th, 2012, 02:26 AM  #5 
Math Team Joined: Apr 2012 Posts: 1,579 Thanks: 22  Re: Probability of Drawing Card
The probability of drawing six straight nonred cards can also be calculated as 35/40 * 34/39 * 33/38 * 32/37 * 31/36 * 30/35 The first fraction is simply the number of nonred cards over all cards in the deck and each successive one is number of nonred cards remaining over all cards remaining given that so and so many nonred cards have been previously drawn. These fractions can then be simplified as: (2*11*17*31)/(3*13*19*37), which give the same answer for the probabity of six nonred cards the others gave. And then you subtract from 1 as they did. The only advantage to doing it this way is that it is more easily hand calculable, as you compute the reduced fraction immediately. 

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