Using Binominal Distribution in Biostatistics, please answer both A and B. Thanks.

Suppose 15% of the female population is left handed. With a random group of 14 people from females, what is:

A. The probability that exactly two of the 14 is left-handed?

Suppose 15% of the female population is left handed. With a random group of 14 people from females, what is:

B. The probability that at least two of the 14 are left-handed?

For A, this is what I got.

q = 1 - 0.15 = 0.85

p = .15

n = 14

k = 2

14! / (2! x 12!) x 0.15^2 x 0.85^12 = .000012