Question 1200581
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A company has a 20-person grievance committee. When a grievance is filed, three of these
20 people are chosen at random to serve on a hearing panel. Suppose that two {{{highlight(cross(committees))}}} <U>panels</U> are
formed. What is the probability that these {{{highlight(cross(committees))}}} <U>panels</U> have at least one member in common?
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<pre>
First  panel can be formed in  {{{C[20]^3}}} = {{{(20*19*18)/(1*2*3)}}} = 1140 different ways.

Second panel can be formed in  {{{C[20]^3}}} = {{{(20*19*18)/(1*2*3)}}} = 1140 different ways.


Two panels can be formed in  {{{C[20]^3*C[20]^3}}} = {{{1140^2}}} different ways.    (1)


Two panels that are not intersect (have empty intersections) can be formed in 

    {{{C[20]^3*C[20-3]^3}}} = {{{C[20]^3*C[17]^3}}} = 1140*680 different ways.          (2)


The probability that two panels have no intersection (have empty intersection) 
is the ratio of number (2) to number (1)

    P = {{{(1140*680)/(1140*1140)}}} = {{{680/1140}}} = {{{34/57}}} = 0.5965  (rounded).  


The probability under the problem's question is the COMPLEMENT to it 

    P' = 1 - 0.5965 = 0.4035.    <U>ANSWER</U>
</pre>

Solved.