Question

Suppose a family has three children of different ages. We assume that all combinations of boys and girls are equally likely. (a) Formulate precisely the sample space and probability measure that describes the genders of the three children in the order in which they are born. (b) Suppose we see the parents with two girls. Assuming we have no other information beyond that at least two of the children are girls, what is the probability that the child we have not yet seen is a boy? Expert Answer

Answer 1

Answer:

a) There are 8 possible combinations and each probability is 1/8.

b) The probability that it is a boy given that there are two girls is 3/8

Step-by-step explanation:

a) The sample space is given by:

BBB (3 boys)

BBG (Boy, boy, girl)

BGB (Boy, girl, boy)

BGG (Boy, girl, girl)

GBB (girl, boy, boy)

GBG (girl, boy, girl)

GGB (girl, girl, boy)

GGG (3 girls)

The probability of each combination is the same:

P(BBB)=P(B∩B∩B)=$\frac{1}{2} \frac{1}{2} \frac{1}{2}=\frac{1}{8}$

2) There are three possible combinations in which there are 2 girls and 1 boy:

BGG, GBG, GGB

So the probability is given by:

P(BGG ∪ GBG ∪ GGB)=$\frac{1}{2} \frac{1}{2} \frac{1}{2}+\frac{1}{2} \frac{1}{2} \frac{1}{2}+\frac{1}{2} \frac{1}{2} \frac{1}{2}=\frac{3}{8}$