Suppose 28.5% of US teens eat while driving. Find the probability that among 13 randomly selected teens at least 3 don't eat while driving.

Question

I think this question has to do with the binomial formula.

Answer 1

Yes it is

 

We will use p = 71.5%    

 

and we will find P(X = 0 or X = 1 or X =2) and subtract from 100%  Easier than doing at least 3: 3,4,5,6,7...13

 

 

P(X = 0)   0 don't eat while driving

 

Only 1 way to have X = 0  so we have 1(.715)⁰(.285)¹³ = .0000000818433

 

P (X = 1)  1 doesn't eat while driving

 

There are 13 ways for this to occur so we have  13(.715)(.285)^{12  =}  .0000026692

 

P(X = 2)   2 don't eat while driving

 

There are 78 ways for this to occur so we have   78(.715)²(.285)¹¹  = .000004017909

 

Add all 3 values = .0000429301 = .00429301%

Now subtract from 100% ANSWER: 99.57%