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 January 4th, 2016, 01:43 PM #1 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 606 Thanks: 82 Probability With Marbles Without Replacement There are 17 green marbles and 3 red marbles. When 4 marbles are selected without replacement, what is the probability of every combination of green and red marbles? P(0 green and 4 red) = 0 P(1 green and 3 red) = x P(2 green and 2 red) = y P(3 green and 1 red) = z P(4 green and 0 red) = (17/20)(16/19)(15/18)(14/17) = 28/57 = about 0.4912280702 Obviously x + y + z = 29/57.
 January 4th, 2016, 05:29 PM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,240 Thanks: 884 Given 17 green and 3 red marbles, the probability that the first marble chosen is green is 17/20. In that case there are 16 green and 3 red marbles so the probability the second marble is also green is 16/19. Then there are 15 green and 3 red marbles so the probability the third marbles is also green is 15/18. Then there are 14 green and 3 red marbles so the probability the fourth marble is red is 3/17. The probability the marbles are "green, green, green, red" is (17/20)(16/19)(15/1(3/17). The same kind of argument shows that the probability of three green and one red in any specific order is also (17/20)(16/19)(15/1(3/17). There are 4!/(3!1!)= 4 different orders ("GGGR", "GGRG", "GRGG", and "RGGG") so the probability of "three green, one red" in any order is 4(17/20)(16/19)(15/1(3/17). The others can be done in the same way.
 January 5th, 2016, 05:14 AM #3 Senior Member   Joined: Apr 2015 From: Planet Earth Posts: 129 Thanks: 25 There is a mathematical function called a “combination.” It requires two arguments: the number of marbles you have, N; and the number you draw without replacement, M. It is: comb(N,M)=N!/M!/(N-M)! What it tells you, is the number of ways you accomplish that draw. So when you choose 4 marbles from your bag of 20, there are comb(20,4)=20!/4!/16!=4845 ways to do it. But you are interested in specific arrangements. To get, say, 2 red marbles (of the 3) and 2 green marbles (of the 17), there are comb(3,2)*comb(17,2)=3*136=408 ways. If you do this for each possibility, comb(3,0)*comb(17,4) = 1*2380 = 2380 comb(3,1)*comb(17,3) = 3*680 = 2040 comb(3,2)*comb(17,2) = 3*136 = 408 comb(3,3)*comb(17,1) = 1*17 = 17 As a check, 2380+680+136+17=4845, the number of total possible combinations. So the probabilities are Pr(0 red, 4 green) = 2380/4845 = 28/57 Pr(1 red, 3 green) = 2040/4845 = 8/19 Pr(2 red, 2 green) = 408/4845 = 8/95 Pr(3 red, 1 green) = 17/4845 = 1/285 Thanks from EvanJ

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