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 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
Two dice are rolled. What is the probability that it is not doubles?
To find the probability in each scenario, we can use the concept of probabilities in independent events.
Given:
- The proportion of underweight Americans is 18% or 0.18.
- We are selecting 4 Americans randomly.
a. Probability that all 4 Americans will be underweight:
Since each selection is independent, we can multiply the probabilities together.
P(all underweight) = P(underweight) * P(underweight) * P(underweight) * P(underweight)
= 0.18 * 0.18 * 0.18 * 0.18
b. Probability that none of the 4 Americans will be underweight:
This is the complement of all Americans being underweight. So, we subtract the probability of all underweight from 1.
P(none underweight) = 1 - P(all underweight)
c. Probability that at least one of the 4 Americans will be underweight:
This is the complement of none of the Americans being underweight. So, we subtract the probability of none underweight from 1.
P(at least one underweight) = 1 - P(none underweight)