SOLUTION: A bag contains 16 fake diamonds and 8 real diamonds that are identically shaped. If 5 diamonds are randomly selected from the bag, what is the probability that 2 are fake and 3 are

Question 1075613: A bag contains 16 fake diamonds and 8 real diamonds that are identically shaped. If 5 diamonds are randomly selected from the bag, what is the probability that 2 are fake and 3 are real?
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
The denominator is 24C5
The numerator is 16C2*8C3; the numbers in front of the C add to 24 and behind add to 5.
The probability is 120*56/42504=0.1581
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Another way is to look at the probability the first is fake second fake, and other three not fake
(16/24)*(15/23)*(8/22)(7/21)*(6/20).
The denominator never changes, but the numerator can occur with 10 different orderings, FFRRR RRFFF etc. It is 5C2 or 5C3, both of which are 10.
The above product is 0.01581, and multiplied by 10 gives 0.1581.
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