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

1 Answer
Oct 19, 2015

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

Explanation:

f (x) = 7C_x7Cx )(0.6)^x(0.6)x (0.4)^(7-x)(0.4)7x , x = 0, 1, ... 7 =>
X follows Binomial Distribution with parameters p = 0.6 and n=7.
7C_47C4 = 7C_37C3 = 7xx× 6xx5×5/ 1xx2×2 xx3×3 = 7xx5×5 = 35
7C_57C5= 7C_27C2 = 7xx6×6 /1xx2×2 = 21, 7C_67C6 = 7C_17C1 = 7 and
7C_77C7 = 1.