Question 41810: From a group of 6 men and 4 women, a committee of 3 is to be selected at random. Find P(at least 2 women).
Answer by fractalier(6550)   (Show Source):
You can put this solution on YOUR website! We can add up the possibilities...
start with having all three women...how many ways can we choose that?
Well 4 chosen three at a time is C(4,3) = 4 ways
Now, let's look at just two women...how many ways can we choose them? Well, C(4,2) = 6 ways...but for each of these ways, we can have any of six men, so there's a total of 36 ways this can happen...
Thus the grand total is 40 different ways.
Let me complete the problem...
How many total ways are there altogether? Well if we complete the ananlysis by looking at C(4,1)*C(6,2), we get
4*15 = 60 one woman committees and
C(6,3) = 20 no woman committees,
so there are a grand total of 120 possible committees, and thus
P(at least two women) = 40/120 = 1/3
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