An experiment consists of recording the sex composition, in order of their births, of a four-child family in which the children were born at different times. assuming that a boy is equally likely as a girl to be born into a family, what is the probability that a four-child family chosen at random will have the following?

(a) three boys and a girl in the family

(b) a youngest child in the family who is a girl

(c) an oldest child and a youngest child in the family who are both boys

a) There is a 25% probability that a four-child family chosen at random will have three boys and a girl in the family.

b) There is a 50% probability of the youngest child being a girl.

c) There is a 25% probability that the oldest child and the youngest child in the family are both boys

Step-by-step explanation:

Probability:

What you want to happen is the desired outcome.

Everything that can happen iis the total outcomes.

The probability is the division of the number of possible outcomes by the number of total outcomes.

In our problem, there is:

-50% probability of giving birth to a boy.

-50% probability of giving birth to a girl.

There are the following babies:

B1 - B2 - B3 - B4

(a) Three boys and a girl in the family:

There is a 50% probability that each children is a boy and 50% that each children is a girl.

We want three boys and a girl, so we have to permutate these probabilities. It is a permutation of 4 elements(all the children), with 4(boys) and 1(girls) repetitions. So:

There is a 25% probability that a four-child family chosen at random will have three boys and a girl in the family.

(b) A youngest child in the family who is a girl

The probabilities are independent from each other. This is just the probability of B4 being a girl, that is 50%.

(c) An oldest child and a youngest child in the family who are both boys

The probabilities of B1 being a boy is 50% and of B4 being a boy is also 50%. So:

There is a 25% probability that the oldest child and the youngest child in the family are both boys