If a couple were planning to have three children, the sample space summarizing the gender outcomes would be: bbb, bbg, bgb, bgg, gbb, gbg, ggb, ggg. a. construct a similar sample space for the possible weightweight outcomes (using o for overweight and u for underweighto for overweight and u for underweight) of two children. b. assuming that the outcomes listed in part (a) were equally likely, find the probability of getting two underweightunderweight children. c. find the probability of getting exactly one overweightoverweight

There are two choices for each child: overweight (o) or underweight (u). So if the first child is o the next can be o or u. If the first is u the second can be o or u. This gives four possibilities. Here the first child is the letter noted first and the second is the one listed second:
OO
OU
UO
UU

There are 4 outcomes and if each is equally likely then the probability of each is 1/4. Thus the probability of UU is 1/4

The probability of one underweight and one over weight is 1/2 because in two of the outcomes listed above there is one O and one U (namely OU and UO). Since there are 4 outcomes the probability is 2/4 = 1/2

Ithink ghe answer to this question is b just trying to others