A total of 28 percent of American males smoke cigarettes, 7 % smoke cigars, .05 smoke cigars and cigarettes. How many smoke cigars but not cigarettes? I was able to find out how many smoke neither cigars nor cigarettes as:
$P(E \cup F)$ is the event that someone smokes a cigar OR a cigarette
$P(E \cup F)^c$ is the event someone smokes neither
$(P \cup F)^c = 1-( P(E) +P(E) - P(EF))= .70$
The probability someone smokes cigars but not cigarettes could be $P(E \cap F^C)$ if I let E denote the event that someone smokes cigar and let F denote the event that a person smokes a cigarette, thus:
$P(E \cap F^C) =....$ I have no clue, this chapter two material over mutually inclusive-exclusive events and their formula and sample spaces with equally likely outcomes and the probability axioms 1-3. So if someone could explain the answer in those terms it would be great. So no Bayes formula although I think it could apply here, I would mind a side note or second part using Bayes method.
 A: Letting $A$ denote the percentage of cigarette smokers and $B$ denote the percentage of cigar smokers, we can write the percent of people who smoke either as
\begin{align*}
P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.28 + 0.07 - 0.05 = 0.30
\end{align*}
So we have
\begin{align*}
P(A^c \cap B) = P(A \cup B) - P(A) = 0.30 - 0.28 = 0.02
\end{align*}
So 2% of people smoke cigars but not cigarettes. You can see the last identity used by drawing a venn diagram as @scott(+1) suggested.
A: Draw a Venn diagram. On the left, you have 28% that smoke cigarettes. On the right, you have the 7% that smoke cigars. The intersection is 5%. Simple arithmetic will tell you the portions of the circles that smoke one but not the other.
A: Let $A$ be the set of Americans, $B$ the set of cigar smokers and $C$ the set of cigarette smokers. Then the question translates to:

Given finite sets $A,B,C$ with $\frac{|C|}{|A|} = .28, \frac{|B|}{|A|} = .07, \frac{|B\cap C|}{|A|} = .0005$, determine $\frac{|B\setminus C|}{|A|}$.

There is absolutely no need to work with probabilities here.
Since for all disjoint finite sets $X,Y$ we have $|X\uplus Y| = |X|+|Y|$, it follows that
$$|B\setminus C| = |B| - |B\cap C| = |A|\cdot(.07-.0005)$$
Thus, $$\frac{|B\setminus C|}{|A|} = .07-.0005 = 0.0695 = 6.95\%$$
Note that the information that 28 percent smoke cigatettes is not required here. Neither are probabilities.
