The probability that a bulb produced in a factory will fuse after 150 days of use is `0.05` Find the probability that out of 5 such bulb :
None will fuse after 150 days of use.

Options

A. `1-(19//20)^(5)`
B. `(19//20)^(5)`
C. `(3//4)^(5)`
D. `90(1//4)^(5)`

Answer

B

Explanation

Here `p=19//20,q=1//20,n=5,r=5.` The required probability is
`=""^(5)C_(3)((19)/(20))^(5)((1)/(20))^(0)=((19)/(20))^(5)`