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I'm given that $P(A)$ = .75, $P(B)$ = .15, and $P(A \cup B)$ = .85. Therefore, $P(A \cap B)$ = .05. Now the question is, what is the probability of A and not B occurring? This may seem like a repeat question, but the only similar questions I could find on this site were trying to find the probability of exactly one occurring. I'm looking for the probability of only A occurring.

My solution was to first find, as above, the probability of only one of the events occurring, which is $P(A \cup B) - P(A \cap B) = .8$. Then, the probability of only A occurring is the probability of A occurring given that only one of the events will occur, or $P(A \mid S)$, where S is the event that only one of A and B occurs. Then the answer is $\frac{P(A \cap S)}{P(S)} = \frac{P(A)}{P(A \cup B) - P(A \cap B)} = \frac{.75}{.8} = .9375$. This doesn't seem correct or simple enough. Any advice is appreciated.

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    $A\cap B$ is a sub-event of $A$, so $P(A\cap B) ≤ P(A)$. Try drawing a diagram...I think that will suggest the right answer. – 2017-02-08

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I'm not sure you need to go so far into this quagmire.

Simply note that $P(A) = P(A \cap B) + P(A \cap B^c)$, since $A \cap B$ and $A \cap B^c$ are mutually exclusive events, and their union is $A$.

Hence, $P(A) - P(A \cap B) = 0.7$ is the answer.

Naturally, you would have realized that $0.9375$ was the wrong answer. Of course, the probability of only $A$ occuring cannot be greater than the probability of it occuring! (which was the case earlier).

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    $A\cap B$ and $A\cap B^c$ are *disjoint* events – 2017-02-08
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    Aah. I am a fool. Thank you for the answer. Out of curiosity, is there a specific reason the method above gives the wrong answer? – 2017-02-08
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    @WyattGregory Yes, it is based on a faulty premise. See my answer for more detail. – 2017-02-08
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    @spaceisdarkgreen Thank you for pointing out the typing error. They are mutually exclusive events, since their intersection is $\phi$. – 2017-02-08
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Does this graphic representing the situation help?

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Then, the probability of only A occurring is the probability of A occurring given that only one of the events will occur

No.   No, it is not.

$$\underbrace{\mathsf P(A\cap B^\complement)}_{\text{what you want}} = \underbrace{\mathsf P(A\mid A\oplus B)}_{\small\text{what you are talking about}}\!\!\!\!\cdot\mathsf P(A\oplus B)$$

NB: $A\oplus B = (A\cup B)\cap(A\cap B)^\complement)$

It is best to find it direct from the Law of Total Probability and the principle of inclusion and exclusion: $$\begin{align}\mathsf P(A)&=\mathsf P(A\cap B)+\mathsf P(A\cap B^\complement)\\ \mathsf P(A\cup B)&=\mathsf P(A)+\mathsf P(B)-\mathsf P(A\cap B) \\ \hline \therefore\quad \mathsf P(A\cap B^\complement)&=\mathsf P(A\cup B)-\mathsf P(B)\end{align}$$

Which may also be found by contemplating a Venn diagram.