# math,pre statistics

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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

• math,pre statistics -

U is underweight, N is not-underweight
a) so you want UUUU = (.18)(.18)(.18)(.18)
= .18^4 = .00105

b) wouldn't that be all not-underweight ?
or (.82)^4 = .45212

c) at least one underweight implies
1 - Prob(all non-overweight)
= 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

• math,pre statistics -

Two dice are rolled. What is the probability that it is not doubles?

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