In a binomial distribution `B(n , p=1/4)` , if the probability of at least one success is greater than or equal to `9/(10)` , then n is greater than (1) `1/((log)_(10)^4-(log)_(10)^3)` (2) `1/((log)_(10)^4+(log)_(10)^3)` (3) `9/((log)_(10)^4-(log)_(10)^3)` (4) `4/((log)_(10)^4-(log)_(10)^3)`

Options

A. `(1)/(log_(10)4-log_(10)3)`
B. `(1)/(log_(10)4+log_(10)3)`
C. `(9)/(log_(10)4-log_(10)3)`
D. `(4)/(log_(10)4-log_(10)3)`

Answer

A

Explanation

(a) According to the condition, `1-((3)/(4))^(n) ge (9)/(10)`
`rArr " " ((3)/(4))^(n) le 1-(9)/(10)=(1)/(10)`
`rArr " " ((4)/(3))^(n) ge 10`
`rArr n[log_(10)4-log_(10)3] ge log_(10)10=1 rArr n ge (1)/(log_(10)4-log_(10)3)`