+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)
+128475 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
