A bag contain three black balls,four white balls and five red balls.Three balls are removed without replacement.What is the probability of obtaining
a)one of each colour
b)at least two red balls?

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  1. So you want BWR
    that particular order's prob = (3/12)(4/11)(5/10) = 1/22
    but the BWR can be arranged in 3! or 6 ways.
    so Prob(your event) = 6/22 = 3/11

    b) could RRB or RRW

    prob(RRB in that order) = (5/12)(4/11)(3/10) = 1/22
    we it could have been RRB, RBR, BRR, so
    prob(of RRB in any order) = 3/22
    do the same for RRW, then add them up

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