in a health study, 18% of americans are underweight. If 4 americans are randomly selected, what is the probability that at least one will be underweight?
prob(underweight) = .18
prob(NOT underweight) = .82
Prob(at least one underweight)
= 1 - prob(all 4 not underweight)
= 1 - .82^4
= .5478..
or appr 55%
here is a more detailed version of your question back from 2009
http://www.jiskha.com/display.cgi?id=1233928692
Thank you so much for taking the time to help me.
To find the probability that at least one out of the four randomly selected Americans will be underweight, you can use the concept of complementary probability.
The complementary probability states that the probability of an event occurring is equal to 1 minus the probability of the event not occurring.
In this case, the event of interest is at least one American being underweight, which is the opposite of none of the Americans being underweight.
To calculate the probability of none of the four Americans being underweight, you need to find the probability that each individual is not underweight and then multiply these probabilities together.
The probability that one American is not underweight is given by (1 - 0.18) = 0.82.
Since the individuals are randomly selected, their probabilities of not being underweight are independent events. Therefore, to calculate the probability that none of the four Americans are underweight, you raise 0.82 to the power of 4 (since there are four individuals).
So, the probability of none of the four Americans being underweight is 0.82^4 ≈ 0.4165.
To find the probability of at least one American being underweight, subtract the probability of none of the Americans being underweight from 1:
1 - 0.4165 ≈ 0.5835.
Therefore, the probability that at least one out of the four randomly selected Americans will be underweight is approximately 0.5835, or 58.35%.