Suppose that $P(X)=a$ and $P(Y)=b$, what is $P(X-Y)$? Suppose that we have:
 $$P(X)=a ~and ~ P(Y)=b$$ 
Now the question is : what is $P(X-Y)$? suppose that X and Y are independent.
$$\newcommand{\d}[1]{{\rm #1}}
\d P(\d A-\d B)=\d P(\d A)-\d P(\d A\d B)=\d P(\d A)-\d P(\d A)\d P(\d B)=a-ab$$
Since for independant events: $\d P(\d A\d B)=\d P(\d A)\d P(\d B)$
Assuming X and Y to be two sets of favourable events....

$P(X) = a$ 
$P(Y)= b$

We are given that X and Y are independent.

Now,

X-Y defines the set of favourable events which belongs to the set X but does not belong to the set Y.

So, we can write $X-Y = X\cap Y^c$ 

;where $Y^c$ defines the compliment of Y.

We already know from the question that $X$ and $Y$ are independent.

Also, without loss of generality $Y$ and $Y^c$ are also independent.

This leads us to conclude that $X$ and $Y^c$ are also independent.

So,

$P(X-Y) = P(X \cap Y^c) = P(X) * P(Y^c)= P(X)*(1- P(Y))=a(1-b) $  

The result is just as everyone before me has concluded.

QED