SOLUTION: If 4 marbles are drawn at random all at once from a bag containing 8 white and 6 black marbles, what is the probability that 2 will be white and 2 will be black?

Question 786752: If 4 marbles are drawn at random all at once
from a bag containing 8 white and 6 black
marbles, what is the probability that 2 will be
white and 2 will be black?
Answer by DrBeeee(684) (Show Source): You can put this solution on YOUR website!
You have 14 marbles, 8 black, b, and 6 white, w.
The probability of selecting a black one is
(1) P(b) = 8/14
Then, without replacement of that black marble (assumed to happen), the probability of selecting another black one is the product of the first probability and the second as given by
(2) P(b,b) = (8/14)*(7/13), note that there is one less black one to choose from and only 13 marbles in the bag because we did not replace the first black. Note also that we assume that the first event happened, i.e. we pulled out a black marble.
Now we want to select a white marble. The probability is the number of white marbles left divided by the total number of marbles left or 6/12. Then to get
(3) P(b,b,w) = (8/14)*(7/13)*(6/12).
Then we want to select another white marble, again without replacement, so we get
(4) P(b,b,w,w) = (8/14)*(7/13)*(6/12)*(5/11).
This is the final answer that simplifies to
(5) P(b,b,w,w) = 10/143
Wait you say! I selected one at a time, not "all at once". My answer is that it doesn't matter; the answer is the same. Not so fast you say; I selected a black, black, white and white, in that "order". Again it doesn't matter; you want 2 black and 2 white. The order doesn't matter. For example, let's calculate the probability of white, black, black, white, and get
(6) P(w,b,b,w) = (6/14)*(8/13)*(7/12)*(5/11)
Compare (6) and (4), they're the same! The numerator is 8*7*6*5 and the denominator is 14*13*12*11, albeit in different order. Ever hear of the commutive law?

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