SOLUTION: 7 men and 3 women are ranked according to their scores on an exam. Assume that no two scores are alike, and that all 10! possible rankings are equally likely. Let X denote the high

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Question 1159882: 7 men and 3 women are ranked according to their scores on an exam. Assume that no two scores are alike, and that all 10! possible rankings are equally likely. Let X denote the highest ranking achieved by a man (so X=1 indicates that a man achieved the highest score on the exam). Find each of the following:
P(X=1)=
P(X=2)=
P(X=3)=
P(X=7)=
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
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7 men and 3 women are ranked according to their scores on an exam. Assume that no two scores are alike,
and that all 10! possible rankings are equally likely. Let X denote the highest ranking achieved by a man
(so X=1 indicates that a man achieved the highest score on the exam). Find each of the following:
(a) P(X=1) =
(b) P(X=2) =
(c) P(X=3) =
(d) P(X=7) =
~~~~~~~~~~~~~~~~~~~~~~~~~~ 

(a)  X=1  means that one of the 7 men is in the first position,
          while the rest 6 men and 3 women are distributed in positions from 2 to 10 in arbitrary ways.

     The number of such possible outcomes is C%5B7%5D%5E1%2A9%21 = 7*9!,
     and to find the probability  P(X=1),  we should relate 7*9!  to  10!.


     It gives  P(X=1) = %287%2A9%21%29%2F10%21 = 7%2F10 = 0.7.     


     It is the  ANSWER to (a).



(b)  X=2 means that one of the 3 women is in the first position and one of the 7 men is in the second position,
         while the rest 3-1 = 2 women and 7-1 = 6 men are distributed in positions from 3 to 10 in arbitrary ways.

     The number of such possible outcomes is C%5B3%5D%5E1%2AC%5B7%5D%5E1%2A%282%2B6%29%21 = 3*7*8!,
     and to find the probability  P(X=2),  we should relate  3*7*8!  to  10!.


     It gives  P(X=2) = %283%2A7%2A8%21%29%2F10%21= %283%2A7%29%2F%289%2A10%29 = 7%2F%283%2A10%29 = 7%2F30.


     It is the ANSWER to (b).



(c)  X=3 means that two of the 3 women are in the first and in the second positions and one of the 7 men 
         is in the third position,
         while the rest 3-2 = 1 women and 7-1 = 6 men are distributed in positions from 3 to 10 in arbitrary ways.

     The number of such possible outcomes is  C%5B3%5D%5E2%2AC%5B7%5D%5E1%2A%286%2B1%29%21 = %28%283%2A2%29%2F2%29%2A7%2A7%21 = 3*7*7!,
     and to find the probability  P(X=3),  we should relate  3*7*7!  to  10!.


     It gives  P(X=3) = %283%2A7%2A7%21%29%2F10%21= %283%2A7%29%2F%288%2A9%2A10%29 = 7%2F%288%2A3%2A10%29 = 7%2F240.


     It is the ANSWER to (c).



(d)  X=7  means that some man is in the 7th positions and there no men in positions from 1 to 6.

          But this configuration is not possible (which is obvious), so

              P(X=7) = 0.


     It is the ANSWER to (d).

Thus, all the questions are answered, and the problem is solved completely.