please help me its a statistic question

Question

It has been reported that 60% of all internet users have been attacked by viruses. If 16 people are randomly surveyed, find the following probabilities:

a. At least 6 have been attacked

b. No more than 4 have been attacked

c. Less than 13 have been attacked

Answer 1

Binomial p=0.6, N=16

 

a)  Prob (X>=6) = 1 - Prob(X<=5)

 

                        = 1 - SUM    (16 choose k) (0.6)^k ( 0.4)^(16-k)    ,  k=0,1,2,3,4,5

 

                        = 0.980858 = 98.0858%

 

b) Prob(X<-4) = SUM (16 choose k) (0.6)^k ( 0.4)^(16-k) , k=0,1,2,3,4 

 

                      = 0.004896 = 0.4896 % < 1/2 %

 

 

c) Prob (X <= 12) = SUM (16 choose k) (0.6)^k ( 0.4)^(16-k) , k=0,1,2,3,4,5,6,7,8,9,10,11,12

 

                        = 0.934853258 = 93.4853258%