a) Since, the historical probability of eggs getting damaged is 15% (that is, about 15 eggs out of 100 are damaged), a random sample of 20 eggs will have 3 (= 20 x 15/100) eggs that are damaged.

b) For a probability of 3 or more damaged eggs, we have to subtract the probabilities of getting 2 damaged eggs, 1 damaged egg and 0 damaged egg from 1.

That is, P(x `>=` 3) = 1 - P (x `>=` 2) - P(x `>=` 1) - P (x = 0)

where x is the number of damaged eggs in the set of 20 sampled eggs.

thus, P (x `>=` 3) = 1 - C (20, 2) x 0.15^2 x 0.85^(20-2) - C (20, 1) x 0.15^1 x 0.85^(20-1) - C (20, 0) x 0.15^0 x 0.85^(20-0) = **0.5951**.

Thus, there is a probability of 0.5951 (or 59.51%) that Johnny Ray will find 3 or more damaged eggs in a sample of 20.

Hope this helps.