Hi,

 9 students in a class: 7 boys and 2 girls.

picks a group of 3 at random

P(all boys) =  

Answer by KOlszewski(2)   (Show Source): You can put this solution on YOUR website!
  

Answer by ikleyn(52786)   (Show Source): You can put this solution on YOUR website!
 .


In this problem, the space of events is the set of all possible different groups of 3 persons that can be formed of 7+2 = 9 students.

The number of such groups is   =  = 84.

It is the number of events in the space of events.


Of them, the favorable set is the set of all different groups that can be formed of 7 boys.

The number of such groups is   =  = 35.


The probability under the problem's question is the ratio of the numer of favorable events (35) to the number of all possible outcomes (84)


    P =  =  = 0.4167 = 41.67%  (approximately).    ANSWER






There is another way to get the SAME answer.





The probability that the first selected student is a boy is  .

If the first selected student is a boy, then the probability that the second selected student is a boy is  .

If the first and the second selected students are boys, then the probability that the third selected student is a boy is  .

Thus the probability under the problem's question is


     = 0.4167 = 41.67%.




Solved.





 

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