Question 1119068: . An apartment building has 20 two-bedroom apartments and 8 three-bedroom apartments. If 4 apartments are chosen at random to be redecorated, what are the probabilities that
a) All will be two-bedroom apartments ?
(b) 2 will be two-bedroom apartments and 2 will be three-bedroom apartments ?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The denominator is 28C4, choosing 4 apartments at random from 28.
The numerator is 20C4, choosing only the two bedroom apartments. The other part of the numerator would be the missing numbers 8C0, but 8C0=1, so it can be ignored.
20C4/28C4=0.2366.
The second will have numerator 20C2*8C2 with the same denominator. Notice that the pre-Cs in the numerator add up to 28 and the post-Cs add up to 4.
probability is 0.2598.
Can check by looking at the other three possibilities, 0,1,3 are two bedroom.
Those probabilities are 0.0547 with numerator 20C1*8C3
and 0.0034 with numerator 20C0*8C4
0.4454 with numerator 20C3*8C1.
Those add to 1.
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