When sampling with replacement you have a binomial distribution for the

number of sucesses, so here the number of redsocks has binomial distribution

B(3,4/9)

so for (i) you want b(2,3,4/9), and for (ii) b(2,3,4/9)+b(3,3,4/9)

Here the probability of two red socks in three choices is(b) Repeat part (a) without replacement

p(n=2)=p(rrb)+p(rbr)+p(brr)

where rrb denotes the first two socks are red and the third is blue, etc.

So:

p(n=2)=(4/9)(3/8)(5/7)+(4/9)(5/8)(3/7)+(5/9)(4/8)(3/7)

and:

p(n>=2)=p(n=2)+p(n=3)

RonL