I am having difficulty with this problem: Consider a group with 4 potential dono

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I am having difficulty with this problem:

Consider a group with 4 potential donors: A,B,C,D - of whom only Ahas type O+ blood. Four samples, one from each individual, will betyped in random order until an O+ individual is identified. LetY={number of typings necessary to identify an O+ individual}.Compute pmf of Y and then answer- what is the probability that morethan two but less than or equal to four typings arenecessary? I have computed my pmf values, and found them all to be 0.25for Y=1,2,3,4 For example, Let O=O+, N=not O+ P(Y=1)=P(O) = 1/4 =0.25 P(Y=2)=P(NO)=(P1,1*P1,3)/P2,4=(1*3)/12 = 0.25 ... I know that the pmf values should add to 1, which they do but,should they all be the same? I tried to follow a book example problem, which was for 5donors where A,B were O+ and B,C,D were not on page 93 of Devore,7th Edition. Any help will be greatly appreciated! Thanks :) I am having difficulty with this problem:

Consider a group with 4 potential donors: A,B,C,D - of whom only Ahas type O+ blood. Four samples, one from each individual, will betyped in random order until an O+ individual is identified. LetY={number of typings necessary to identify an O+ individual}.Compute pmf of Y and then answer- what is the probability that morethan two but less than or equal to four typings arenecessary? I have computed my pmf values, and found them all to be 0.25for Y=1,2,3,4 For example, Let O=O+, N=not O+ P(Y=1)=P(O) = 1/4 =0.25 P(Y=2)=P(NO)=(P1,1*P1,3)/P2,4=(1*3)/12 = 0.25 ... I know that the pmf values should add to 1, which they do but,should they all be the same? I tried to follow a book example problem, which was for 5donors where A,B were O+ and B,C,D were not on page 93 of Devore,7th Edition. Any help will be greatly appreciated! Thanks :)

Explanation / Answer

They are all the same first (1/4) = .25 second (3/4)(1/3) = .25 = probability that it was not thefirst one times the probability that it was the second one first (1/4) = .25 second (3/4)(1/3) = .25 = probability that it was not thefirst one times the probability that it was the second one third (3/4)(2/3)(1/2) = .25 fourth (3/4)(2/3(1/2) = .25 what is the probability that more than two but less than orequal to four typings are necessary? that would mean it is the either third or the fourth try, soyou add those two probabilities .25 + .25 = .5

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