3 beer drinkers (1,2,3) are to rank 4 different brands of beer( A,B,C,D). each drinker ranks the four beers as 1 for best and 4 for worst. after all the beers are ranked, the ranks of each brand of beer are summed.
what is the probability that some beer will have a total rank of 4 or less?
the sample points= (4!)^3
but im not sure how to count the individual rankings for the qns...
P( sum =3 ) = (1/4 )^3
P( sum =4) =
There are only very few ways for beer A, say, to have a total rank of 4 or less: either (1,1,1), (2,1,1), (1,2,1) or (1,1,2) (I'm listing the rankings given to beer A by the three drinkers). So that the probability that beer A has a total rank of 4 or less is... (the answer is not 4/(4!)^3...)
Originally Posted by alexandrabel90
And you can notice that it is impossible for two beers to have simultaneously a total rank less than 4 (each of these beers needs to have at least two '1'...). Therefore, we have disjoint events, so that the probability that "some" beer has a total rank less than 4 is four times the previous probability (either beer A, or beer B, or beer C, or beer D has rank less than 4, and these events are disjoint and have same probability).