The probability that a students fail a certain exam has been estimated to be 0.05.

The probability that at least three scripts will be needed until the professor who marks the scripts finds the second script to be a fail is

(a) 1 − (0.05)2(0.95)

(b) 1 − (0.05)2

(c) (0.05)2 + (0.05)2(0.95)

(d) (0.95)2

(e) (0.95)2(0.05)