Question

Please can some one help?
A person is selected at random from a population that has the following characteristics: 65% of the people are men; the others are women 12% of the men are smokers 7% of the women are smokers
i) Find the chance that the selected person is a man or a non-smoker
P(man-non-smoker)= 0.65*(1-0.12)=0.65*0.88=0.572
P(women-nonsmoker)=0.35*(1-0.07)=0.35*0.93=0.3255
P(non-smoker)=(p(man non-smoker + women non-smoker)=0.572+0.3255=0.8975
P(man or a non-smoker)=P(man) + P(non-smoker) - P(man of non-smoker) = (0.65+0.8975)-(0.572)=0.9755 inclusion method ok???

Answer 1

I believe that is correct. Since we have:
\[
|A\cup B|=|A|+|B|-|A\cap B|
\]For \(|A|\) is the cardinality of set \(A\).

Answer 2

Thanks

Answer 3

Infinity does it seem ok with you too?

Answer 4

Alternatively, you may also go this way:
P(man or non-smoker)=P(man)+P(woman and nonsmoker)=0.65+0.35×0.93=0.9755

Answer 5

That works, too, I believe, since \(P(\text{man})\) and \(P(\text{non-smoking woman})\) are mutually-exclusive events, as they share no common points. (Note that, for the set of samples \(M\) of the subject being a man, \(M^c=W\), where \(W\) is the sample-set of the subject being a woman).

Answer 6

Thanks kropot72 You are great!

Answer 7

Thanks lolwolf and kropot72Please adjust the number -P(0.65*(1-0.12))=-(O.65*0.88)=(0.572)=P(man non-smoker)= the final Answers=0.9755:)