Stat 100

[BarTopDancer]

Thanks guys! I'm still lost on B (I had a feeling it was wrong since it was greater than 0), but I think I fixed C and D.

b. If three Americans are randomly selected, what is the probability that all three regularly engage in exercise? Set up the calculation.

.80 x .20 = .16
.16 x 3 = .48

c. What assumption is being made in part b above?

The assumption being made in part B is that if you remove one person from the equation it will not change the outcome of the probability.

d. If two Americans are randomly selected, what is the probability that the first American regularly engages in exercise while the second American does not regularly engage in exercise? Set up the calculation.

.80 x .20 = .16
1 - .20 = .80