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

a) The sample space is {BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG} with each outcome having a probability of 1/8. b) The probability that the child we have not yet seen is a boy is 4/5.

Step-by-step explanation:

a) To formulate the sample space and probability measure, we need to consider the possible outcomes for the genders of the three children in the order they are born. Let's denote a boy as B and a girl as G. The sample space will be {BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG}, where the first letter represents the gender of the oldest child, the second letter represents the gender of the second oldest child, and the third letter represents the gender of the youngest child. Each outcome in the sample space is equally likely, so the probability measure for each outcome is 1/8.

b) Given that we have seen the parents with two girls, we can eliminate the outcomes {BBB, BBG, GBB} from the sample space because these outcomes do not have at least two girls. The remaining outcomes are {BGB, BGG, GBG, GGB, GGG}. Out of these five outcomes, four have at least one boy. Therefore, the probability that the child we have not yet seen is a boy is 4/5.

Answer 2

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)=
`(1)/(2) (1)/(2) (1)/(2)=(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)=
`(1)/(2) (1)/(2) (1)/(2)+(1)/(2) (1)/(2) (1)/(2)+(1)/(2) (1)/(2) (1)/(2)=(3)/(8)`