Question 1166019
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the denominator is the number of ways to choose 3 from (6+4+7) = 17, which is  {{{C[17]^3}}} = {{{(17*16*15)/(1*2*3)}}} = 680.


all teachers is  {{{C[6]^3}}} = {{{(6*5*4)/(1*2*3)}}} = 20 ways.


So the answer to (a) is  P = {{{20/680}}} = 0.029412.




no teachers is  {{{C[4+7]^3}}} = {{{C[11]^3}}} = {{{(11*10*9)/(1*2*3)}}} = 165 ways.


So the answer to (b) is  P = {{{165/680}}} = 0.242647.
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