The Canadian Tobacco Monitoring Survey is a national survey administered by Statistics Canada to study smoking trends of Canadians aged $15$ or older. The most recent survey found that $19$% of Canadians aged $15$ years or older smoke on a daily basis. In addition, $22$% of men are smokers, and $16$% of women are smokers.
The most recent census shows that men make up $50$% of the Canadian population, with women making up the remainder.
You randomly pick a Canadian that is $15$ years old or older. What is the probability this person
a) Is male and a smoker?
b) is a women and not a smoker?
c) What percentage of smokers are male?
d) What percentage of smokers are women?
I think I've got $1/2$ of a probability table figured out,
P(Male) = $0.5$
P(Male complement) = $0.5$,
P(Smoker) = $0.19$
P(Smoker Complement) = $0.81$
I'm just stuck on how to fill in the middle. Would appreciate any help!
Best Answer
Let $M$ denote the set of males; let $S$ denote the set of smokers. Then the probability we seek is $$P(M \cap S) = P(S \mid M)P(M) = 0.22 \cdot 0.5 = 0.11$$ as you found.
Let $F$ denote the set of females; let $S$ denote the set of smokers. Then the probability we seek is $$P(F \cap S^C) = P(S^C \mid F)P(F) = (1 - 0.16)(0.5) = 0.84 \cdot 0.5 = 0.42$$ again as you found.
We need to find the probability that a person is male given that the person is a smoker.
\begin{align*} P(S \mid M) & = \frac{P(M \cap S)}{P(S)}\\ & = \frac{P(M \cap S)}{P(M \cap S) + P(F \cap S)}\\ & = \frac{P(S \mid M)P(M)}{P(S \mid M)P(M) + P(F \mid M)P(F)}\\ & = \frac{0.22 \cdot 0.5}{0.22 \cdot 0.5 + 0.16 \cdot 0.5}\\ & = \frac{0.11}{0.11 + 0.08}\\ & = \frac{0.11}{0.19}\\ & = \frac{11}{19} \end{align*} which you can convert into a percentage.
Subtract the percentage of smokers who are male from $100\%$.