Answer
a. $\frac{1}{8}$
b. $\frac{1}{4}$
c. $\frac{3}{4}$
d. $\frac{3}{4}$
Work Step by Step
Each child can be a boy or a girl, the total outcome is $N=2\times2\times2=8$
a. All boys
There is only 1 favorable outcome, the probability is $p=\frac{1}{8}$
b. All girls or all boys
There are 2 favorable outcome, the probability is $p=\frac{2}{8}=\frac{1}{4}$
c. Exactly two boys or two girls
There are 6 (bbg,bgb,gbb,ggb,gbg,bgg) favorable outcome, the probability is $p=\frac{6}{8}=\frac{3}{4}$
d. At least one child of each gender
This is the same as c. so $p=\frac{3}{4}$
