The probabilities of a student getting I, II and III division in an examination are respectively $\frac{1}{10}, \frac{3}{5}$ and $\frac{1}{4}.$ The probability that the student fail in the examination is

none of these

A denote the event getting I;

B denote the event getting II;

C denote the event getting III; and

D denote the event getting fail.

Obviously, these four events are mutually exclusive and exhaustive, therefore

P(A)+ P(B)+ P(C)+ P(D) = 1 ⇒ P(D) = 1 − 0.95 = 0.05.