SOLUTION: There are 4 students attending a test and 7 teachers available for checking the tests. Each student's test is checked by exactly one of these teachers, with each teacher being equa

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Question 1135588: There are 4 students attending a test and 7 teachers available for checking the tests. Each student's test is checked by exactly one of these teachers, with each teacher being equally likely to be the one who checks the test. The probability that all the 4 papers are checked by exactly 3 teachers is k(6/7)^2.
What is the value of k?
(Sorry for leaving out the last part)

Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
.

The formulation of this post seems to be tricky, so we need re-formulate it in clean, clear and simple form.


The 4 papers can be checked by any team of 4 teachers; by any team of 3 teachers; by any team of 2 teachers; 
and, finally, by any one of 7 teachers.


So we need to calculate the sum 


    sum = C%5B7%5D%5E4 + C%5B7%5D%5E3 + C%5B7%5D%5E2 + C%5B7%5D%5E1     (which is the number of elements in the space of events)


and then calculate the probability as the ratio  P = C%5B7%5D%5E3%2Fsum,   where  C%5B7%5D%5E3 is the number of favorable events.


In this way we will get the answer.


    C%5B7%5D%5E4 = %287%2A6%2A5%29%2F%281%2A2%2A3%29 = 35;

    C%5B7%5D%5E3 = %287%2A6%2A5%29%2F%281%2A2%2A3%29 = 35;

    C%5B7%5D%5E2 = %287%2A6%29%2F%281%2A2%29   = 21;

    C%5B7%5D%5E1           = 7.


So,  sum = 35 + 35 + 21 + 7 = 98.


Therefore,  P = C%5B7%5D%5E3%2Fsum = 35%2F98 = 5%2F14.


Since we are given that   P = 5%2F14 = k%2A%286%2F7%29%5E2,  it implies


    k = %285%2F14%29%2A%287%2F6%29%5E2 = %285%2A7%2A7%29%2F%2814%2A6%2A6%29 = %285%2A7%29%2F%282%2A6%2A6%29 = 35%2F72.    ANSWER


ANSWER.  k = 35%2F72.

SOLVED.