Question 1135588
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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[7]^4}}} + {{{C[7]^3}}} + {{{C[7]^2}}} + {{{C[7]^1}}}     (which is the number of elements in the space of events)


and then calculate the probability as the ratio  P = {{{C[7]^3/sum}}},   where  {{{C[7]^3}}} is the number of favorable events.


In this way we will get the answer.


    {{{C[7]^4}}} = {{{(7*6*5)/(1*2*3)}}} = 35;

    {{{C[7]^3}}} = {{{(7*6*5)/(1*2*3)}}} = 35;

    {{{C[7]^2}}} = {{{(7*6)/(1*2)}}}   = 21;

    {{{C[7]^1}}}           = 7.


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


Therefore,  P = {{{C[7]^3/sum}}} = {{{35/98}}} = {{{5/14}}}.


Since we are given that   P = {{{5/14}}} = {{{k*(6/7)^2}}},  it implies


    k = {{{(5/14)*(7/6)^2}}} = {{{(5*7*7)/(14*6*6)}}} = {{{(5*7)/(2*6*6)}}} = {{{35/72}}}.    <U>ANSWER</U>


<U>ANSWER</U>.  k = {{{35/72}}}.
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SOLVED.