For problem one part b. there are 4 ways the white ball

Can be chosen at least 2 times.

WWW

WWR

WRW

RWW

The combined probability is the sum of the probability of each of these.

So P(www) = 1/4 * 1/4 * 1/4 = 1/64

P(WWR) = P(WRW) = P(RWW) = 1/4 * 1/4 * 3/4 = 3/64

So adding them up gives (3 * 3/64) + 1/64 = 10/64 = 5/32.