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