math,pre statistics
posted by Angela .
In a recent health study it was claimed that 18% of americans are underwieght.If 4 americans randomly selected what is the probability that
a. that all will be underweight
b.none will be underweight
c. at least one will be underweight

U is underweight, N is notunderweight
a) so you want UUUU = (.18)(.18)(.18)(.18)
= .18^4 = .00105
b) wouldn't that be all notunderweight ?
or (.82)^4 = .45212
c) at least one underweight implies
1  Prob(all nonoverweight)
= 1.45212
Since there are only 5 cases possible, I will do each one to show the logic of the above
none underweight = (.82)^4 = .45212
1 underweight = C(4,1)(.18)(.82)^3 = .39698
2 underweight = C(4,2)(.18)^2(.82)^2 = .13071
3 underweight = C(4,3)(.18)^3(.82) = .01913
4 underweight = (.18)^4 = .00105
notice if we total this we get 1 
Two dice are rolled. What is the probability that it is not doubles?
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