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

Question

a)one of each colour
b)at least two red balls?

Answer 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