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
pr(a.)=18/100*18/100*18/100*18/100
I will be happy to check the others. On the last, remember there are four ways the combination can occur
Sorry, on the last, you could have
4 ways to have one underweight
UFFF, FUFF, FFUF, FFFU
Ways to have two U
UUff, UfUf, UffU, FUUf, FUfU, FfUU
Ways to have three U
UUUf, UfUU, FUUU, UUfU
Ways to have four U.
UUUU
To solve these probability questions, we need to use the concept of probability and basic probability calculations. We can calculate the probability for each scenario using the given information.
a. Probability that all four will be underweight:
Given that 18% of Americans are underweight, we can say that the probability of selecting an underweight American is 0.18 (or 18%). Since the selections are independent events (one selection does not affect the probability of the next selection), we can multiply the probabilities together for each selection:
P(all underweight) = 0.18 * 0.18 * 0.18 * 0.18 = 0.000072 or 0.0072%
So, the probability that all four selected Americans will be underweight is 0.000072 or 0.0072%.
b. Probability that none will be underweight:
The probability of selecting a non-underweight American is 1 - 0.18 = 0.82 (or 82%). Again, since the selections are independent events, we can multiply the probabilities together for each selection:
P(none underweight) = 0.82 * 0.82 * 0.82 * 0.82 = 0.4085 or 40.85%
So, the probability that none of the four selected Americans will be underweight is 0.4085 or 40.85%.
c. Probability of at least one underweight:
To find the probability of at least one underweight American, we need to find the complement of the probability that none will be underweight. The complement is equal to 1 minus the probability of the event not occurring. In this case, the probability of none being underweight is already calculated in part b.
P(at least one underweight) = 1 - P(none underweight)
P(at least one underweight) = 1 - 0.4085
P(at least one underweight) = 0.5915 or 59.15%
So, the probability that at least one of the four selected Americans will be underweight is 0.5915 or 59.15%.