For a binomial distribution `B(n, p), np = 4` and varience npq = 4/3. What is the probability `P (x ge 5)` equal to ?

Options

A. `(2//3)^(6)`
B. `(1//3)^(6)`
C. `(1//3)^(6)`
D. `(2^(8)//3^(6))`

Answer

D

Explanation

Given, `np = 4 and npq = 4/3`
`therefore 4q = 4/3 rArr q = 1/3`
`therefore p = 1 - 1/3 = 2/3`
`rArr n = (4 xx 3)/(2) = 6`
Now, `P (X ge 5) = ""^(6)C_(5) (p)^(5)(q)^(1) + ""^(6)C_(6)p^(6)q^(0)`
`= ""^(6)C_(5) (2/3)^(5) (1/3) + ""^(6)C_(6) (2/3)^(6)`
`= (6 xx 32)/(3^(6)) + (64)/(3^(6)) = (256)/(3^(6)) = 2^(8)/3^(6)`