Can anyone solve this obesity statistical problem?

A recent poll states that 35% of all American adults are overweight. If 5 adults are selected at random, what is the probability that at least one adult is overweight?

p(X>0) = 1 -p(0) {prob that more than zero people are overweight}

prob of 0 people having positive condition is 1(.35)^0(.65)^5 = 11.6%

1 - (p(0)) = .88397 or 88.4%

35 in a group of 100 people are statistically obese.
so, out of a group of 5, there will be (35 / 100 * 5 ) obese people; or in other terms, there is/are 1.75 obese people.
The answer to your question in therefore : 100% probable
The only selection of adults that doesn't fulfill "at least one is overweight" is if they are all slim

The probability of all being slim
=65% x 65% x 65% x 65% x 65%........(you get the 65% from 1 - 35%)....They must be overweight or slim, so these add up to 100%

This is approx = 2/3 x 2/3 x 2/3 x 2/3 x 2/3

approx = 32/243 x 100%

approx = 1/7 x 100%

=14% in very round numbers - I'll let you do the arithmetic as I haven't got a calculator

Every other varietion of people fits "at least one overweight

So 1 - the only invalid choice gives you the probability of the valid choices.

Approx 85% probability that at least one is over weight