Question

For binomial distribution X, with n = 7 and p = 0.6, what is P(X>3)?

Answer 1

Pr{X>3} = f(4) + f(5) + f(6) + f(7) = `(0.6) ^4` [35`xx` `(0.4)^3`
+ 21`xx` 0.6`xx` `(0.4)^2` + 7`xx` `(0.6)^2` `xx` 0.4 + `(0.6)^3` ]
=0.710208

Explanation:

f (x) = `7C_x` )`(0.6)^x` `(0.4)^(7-x)` , x = 0, 1, ... 7 `=>`
X follows Binomial Distribution with parameters p = 0.6 and n=7.
`7C_4` =`7C_3` = 7`xx` 6`xx5`/ 1`xx2` `xx3` = 7`xx5` = 35
`7C_5`= `7C_2` = 7`xx6` /1`xx2` = 21, `7C_6` = `7C_1` = 7 and
`7C_7` = 1.