Question 1144390
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            Assuming that the probability to have in a family a boy  (B)  or a girl  (G)  is   {{{1/2}}}



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(A)  The probability that all 3 children are of the same gender =     


         = The probability of  {{{(1/2)*(1/2)*(1/2)}}} = {{{1/8}}}  to have 3 girls 

     PLUS

           the probability of  {{{(1/2)*(1/2)*(1/2)}}} = {{{1/8}}}  to have 3 boys =


     = {{{1/8}}} + {{{1/8}}} = {{{2/8}}} = {{{1/4}}}.    <U>ANSWER</U>




(B)  P = P(BBG) + P(BGB) + P(GBB) + P(GGB) + P(GBG) + P(GGB) = {{{6/8}}} = {{{3/4}}}.     <U>ANSWER</U>



Notice that the probability of the n.B  is the complement to the probability of the n.A, as it should be.


It is an additional argument/check that the solution is correct.
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Solved.