Question 1119078
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Ten students, A, B,..., are in a class. A committee of three is chosen at random to represent the class. Find the
probability that:


<pre>
Of 10 students of the class, you can form  {{{C[10]^3}}} = {{{(10*9*8)/(1*2*3)}}} = 120 triples (triplets), in all.



(a) A belongs to the committee;        You can form  {{{(9*8)/2}}} = 36 such triples, by adding two other of remaining 9

                                       students of the class.

                                       So the probability under the question is  {{{36/120}}} = {{{3/10}}} = 0.3 = 30%



(b) B belongs to the committee;        Same answer (and same solution) as (a).




(c) A and B belong to the committee;   You can form only 8 such triples by adding any of remaining 8 students to A and B.

                                       The answer is  {{{8/120}}} = {{{1/15}}}.



(d) A or B belong to the committee;    Add (a) with (b) and subtract (c):

                                       {{{3/10 + 3/10 - 1/15}}} = {{{9/30 + 9/30 - 2/30}}} = {{{16/30}}} = {{{8/15}}}.
</pre>

Solved.