SOLUTION: A bag contain 6 red, 4 blue, 2 green and 3 yellow marbles. If four marbles are picked at random,what is the probability that one is green, two are blue and one is red?

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Question 892771: A bag contain 6 red, 4 blue, 2 green and 3 yellow marbles.
If four marbles are picked at random,what is the probability that one is green, two are blue and one is red?

Answer by AnlytcPhil(1806)   (Show Source): You can put this solution on YOUR website!
a bag contain 6 red,4 blue,2 green and 3 yellow marbles.
 if four marbles are picked at random,what is the probability that  one is green,two are blue and one is red?

We can pick the 1 green marble any of C(2,1) = 2 ways.
We can pick the 2 blue ones any of C(4,2) = 6 ways.
We can pick the 1 red one any of C(6,1) = 6 ways.

The number of ways we can pick that is 2*6*6 = 72

That's the numerator of the probability.

The denominator of the probability is the number of ways we can pick 
any 4 marbles out of the 6+4+2=12.

That's C(12,4) = 495 ways.

So the probabiloity is 72 successful ways out of 495 possible ways.

Answer  which reduces to .

Edwin


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