+0  
 
0
46
1
avatar+4 

Studies report that approximately 9.2% of men have a type of color blindness that prevents them from distinguishing between red and green. If 22 men are chosen at random, find the following

(a) That exactly 7 have this type of color blindness(I know the answer is 0.22%, but idk how I got it)
 



(b) That at most 5 have this type of color blindness(I know the answer is .09876, but idk how I got it)



 

(c) That atleast 3 have this type of color blindness( I know the answer is 0.3299, but idk how I got it)

AngelRicher  Apr 30, 2018
edited by Guest  Apr 30, 2018
edited by AngelRicher  Apr 30, 2018
Sort: 

1+0 Answers

 #1
avatar+86528 
+1

These are binomial probabilites

 

First one

 

C(22,7) (.092)^7* (.908)^15  = .00236  ≈  .22% 

 

 

Second one  : At  most 5   =   1 - P(7 have colorblindness) - P(6 have colorblindness)  =

 

1 - .00236  - C(22,6)(.092)^6 (.908)^16  = .9879  

 

 

Last one :  At least 3  =  1  - P(0) - P(1) - P(2)  =

 

1 - C(22,0)(.908)^22  - C(22,1)(.092)(.908)^21 - C(22,2)(.092)^2(.908)^20  = .3299

 

 

cool cool cool

CPhill  Apr 30, 2018

20 Online Users

New Privacy Policy (May 2018)
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  Privacy Policy