Question 959340: The participants in a television quiz show are picked from a large pool of applicants with approximately equal numbers of men and women. Among the last 12 participants there have been only 2 women. If participants are picked randomly, what is the probability of getting 2 or fewer women when 12 people are picked?
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Binomial Problem with n = 12 and p = 1/2
My question is I don't get how p=1/2, why is that? What would the value of q be then?
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! This is a binomial probability problem and there is probability(Pr) of (1/2) that person selected will be a women, we make 12 picks and are asked what is the Pr of getting 0, 1, 2 women in the 12 picks, that is
Pr( 0,1,2) = Pr(0) + Pr(1) + Pr(2),
binomial Pr = ( n! / (x!*(n-x)!)*(Pr^x)*((1-Pr)^(n-x))
Pr(0) = ( 12! / (0!*12!)*((1/2)^0)((1/2)^12) = 0.000244141
Pr(1) = ( 12! / (1!*11!))*(1/2)*(1/2)^11 = 0.002929688
Pr(2) = ( 12! / (2!*10!))*(1/2)^2*(1/2)^10 = 0.016113281
Pr( 0,1,2) = 0.000244141 + 0.002929688 + 0.016113281
Pr( 0,1,2) = 0.01928711 which is approx 0.02
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