This is binomial.
part c,
An auditor is erroneous 5% of the time when preparing financial statements. Suppose that he randomly checks 3 entries, find the probability that he makes
a) 0 mistakes
b) 1 mistakes
c) 2 mistakes
d) 3 mistakes
for a) I know it's (0.95)(0.95)(0.95) = 0.857375
for b) 1-0.857375-0.000125-(whatever you get in c)
for c) I would think that it's (0.05)(0.05)(0.95) = 0.002375, but I know that this is incorrect since then solution in the back of the book is 0.007125
for d) I know it's (0.05)(0.05)(0.05) = 0.000125