For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability that the candidate can solve any problem is `(4)/(5)` then the probability that he is unable to solve less than two problem is :

Options

A. `201/5(1/5)^(49)`
B. `54/5(4/5)^(49)`
C. `134/25(1/5)^(48)`
D. `316/25(4/5)^(48)`

Answer

B

Explanation

`P(s)=4/5,P(f)=1/5`
P(o failiure + failure) `=(4/5)^(50)+.^(50)C_(1)xx(4/5)^(49)xx(1/4)=(4/5)^(49)(4/5+10)=(4/5)^(49)(54/5)`