Question 1168580: Two marbles are drawn without replacement from a box with 3 white, 2 green, and 2 red and 1 blue marble: Find the probability: One marble is green and one marble is red
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the box has 3 white, 2 green, 2 red, 1 blue marble.
if you draw 2 marbles from the box, you want to know the probability that one of those marbles is red and one of those marbles is green.
couple of ways to solve this.
one way is to use the combination formula.
the number of ways you can get a set of 2 out of a set of 8 is c(8,2) = 28
the number of ways you can get a set of 1 green out of a set of 2 greens is c(2,1) = 2.
the number of ways you can get a set of 1 red out of a set of 2 reds is c(2,1) = 2.
the probability of getting a red and a green out of a set of 8 is therefore (2*2)/28 = 4/28 = 1/7.
another way is to determine that the probability of getting a green and a red out of a set of 8 that contains 2 greens and 2 reds is:
p(first green and then red) = 2/8 * 2/7 = 4/56.
p(first red and then green) = 2/8 * 2/7 = 4/56.
total probability is p(first green and then red) + p(first red and then green) = 8/56.
simplify to get 1/7.
both ways assume drawing without replacement.
both ways get the same answer.
the answer appears to be 1/7.
the combination formula says c(n,x) = n! / (x! * (n-x)!)
then n = 8 and x = 2, that formula becomes 8! / (2! * 6!)
8! can be shown as (8 * 7 * 6!)
the formula becomes (8 * 7 * 6!) / (2! * 6!)
6! cancels out, so the formula becomes (8 * 7) / (2 * 1)
solve to get c(8,2) = 56/2 = 28.
that was the denominator in the first way to solve the problem.
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