According to recent studies, 57.6% of American citizens are overweight. Suppose that 17% of those who...

According to recent studies, 57.6% of American citizens are overweight. Suppose that 17% of those who are overweight are children. If American citizen is randomly selected, determine the following probabilities:

a) Selected citizen is overweight and a child

b) Selected citizen is not a child given that he/she is overweight

c) Selected citizen is not a child and is not overweight

Expert Solution

P(overweight) = 0.576

P(children | overweight) = 0.17

a) P(overweight and child) = P(children | overweight) * P(overweight)

= 0.17 * 0.576

= 0.0979

b) P(not child | overweight) = 1 - P(children | overweight)

= 1 - 0.17

= 0.83

c) Probability of children who given that they are not overweight is missing. So we can't find P(not child and not overweight)

milcah answered 1 year ago

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