Question 1154150
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In this problem, the probability is the ratio/fraction, whose denominator is the same for all 3 parts, a), b) and c).


This denominator is  {{{C[6]^3}}} = {{{(6*5*4)/(1*2*3)}}} = 20, and it represents the number of triples that can be formed from 6 = 1+2+3 balls.


The numerator is different for each part. It is the number of favorable triples in each case.



(a)  In this case, there is only one favorable triple, consisting of 3 blue balls.


     Thus the probability in this case is  P = {{{1/20}}} = 0.05.



(b)  In this case, favorable triples can be formed from  5 = 2+3 yellow and blue balls.

     The number of favorable triples is, therefore,  {{{C[5]^2}}} = 10.

     Thus the probability in this case is  P = {{{10/20}}} = {{{1/2}}} = 0.5.



(c)  This time the triples contain 2 yellow balls, and we can add EITHER 1 white or any one of 3 blue balls.

     Hence, the number of different favorable triples in this case is  1+3 = 4.


     Therefore,  the probability is  P = {{{4/20}}} = {{{1/5}}} = 0.2.
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Solved.  // All questions are answered.