Question 909190
X~binomial(n=50, p=.07)

A) P[X<3] = P[X<=2] = P[X=0] + P[X=1] + P[X=2] = (50 choose 0)(.07)^0(.93)^50 + (50 choose 1)(.07)^1(.93)^49 + (50 choose 2)(.07)^2(.93)^48 = .3108

B) P[X=2] + P[X=3] + P[X=4] = (50 choose 2)(.07)^2(.93)^48 + (50 choose 3)(.07)^3(.93)^47 + (50 choose 4)(.07)^4(.93)^46 = .6025

C) P[X>5] = 1-P[X<=5]  (using the complement). 

1-(P[X=0] + P[X=1] + (P[X=2] + P[X=3] + P[X=4])+ P[X=5])

1-(P[X=0] + P[X=1] + B + P[X=5])  where B is the answer from part B.

1-((50 choose 0)(.07)^0(.93)^50 + (50 choose 1)(.07)^1(.93)^49 + .6025  + (50 choose 5)(.07)^5(.93)^45) = .1351

Hope this helps!

Devin [swincher4391@yahoo.com]