A family has four children and we assume that births occur with equal frequency for both sexes. Construct a sample space for all possible family arrangements allowing for order of birth to count.

a) How many different arrangements are possible?
b)find P (three are boys)=
c)find P(less than three are girls)=

To construct a sample space for the possible family arrangements, we can use the concept of a binary string. Let's assign "B" to represent a boy and "G" to represent a girl.

The possible arrangements for a family with four children would be:
1) BBBB
2) BBGB
3) BGBB
4) BGGG
5) GBGB
6) GBGG
7) GGBB
8) GGGB
9) GGGG

a) There are 9 different arrangements possible.

b) To find the probability of having three boys, we need to determine the number of favorable outcomes and divide it by the total number of possible outcomes.

Favorable outcomes: There are four ways to have three boys (BBBG, BBGB, BGBB, GBBB).

Total outcomes: There are nine possible arrangements, as we found above.

Therefore, the probability of having three boys is P(three are boys) = favorable outcomes / total outcomes = 4/9.

c) To find the probability of having less than three girls, we need to determine the number of favorable outcomes (0, 1, or 2 girls) and divide it by the total number of possible outcomes.

Favorable outcomes: There are seven ways to have less than three girls (BBBB, BBGB, BGBB, GBGB, GGBB, GGGB, GGGG).

Total outcomes: There are nine possible arrangements, as we found above.

Therefore, the probability of having less than three girls is P(less than three are girls) = favorable outcomes / total outcomes = 7/9.