Answer:
a) X={BBB;BBG:BGG:GGG}
b) P(BBB)=0.125
c) P(BBB)=0.133
Explanation:
The sample space states all the possible values that the random variable can take. In this case, the order does not matter, so the possible combinations for the random variable are:
X={BBB;BBG:BGG:GGG}
The probability p of having a boy is p=0.5, as it is equally likely to have a girl or a boy.
The probability that all 3 children are boys can be calculated multiplying 3 times the probability of having a boy. That is:

In the case that the chance of having a boy is p'=0.51, the probabiltity of having 3 boys become:
