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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)

 Apr 30, 2018
edited by Guest  Apr 30, 2018
edited by AngelRicher  Apr 30, 2018
 #1
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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

 Apr 30, 2018

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