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 similiar sample space for the possible weight outcomes (using o for overweight and u for underweight) of two children. b.) assuming that the outcomes listed in part (a) were equally likely, find the probablity of getting two underweight children. c.) find the probability of getting one overweight child and one underweight child.

(A) oo, ou, uo, uu

(B) 1/4

(C) 1/2

Step-by-step explanation:

(A) When using o for overweight and u for underweight, there are four possible weight outcomes which are; oo, ou, uo, uu

                   The sample space would be: oo, ou, uo, uu

This implies there are 4 possible outcomes.

(B) From the sample space, the event, getting two underweight weight children occurs only once, uu. The probability of getting two underweight children = 1/4

(C)  From the sample space, the event, getting one overweight child and one underweight child occurs twice, ou, uo.

    The probability of getting one overweight child and one underweight child = 2/4 = 1/2

x

step-by-step explanation:

the mistake in the pattern is that :

b. the decimal point was moved one too many place values each time

the correct pattern could be made by :

b. moving the decimal point to the right the same numbers of place values as the exponents