Rolling two dice, what is the probability of getting 6 on either of them, but not both? Rolling two dice, what is the probability of getting 6 on one of them, but not both?

If the first die shows $6$ and the second shows anything but $6$ that can happen $5$ ways (6-1, 6-2, 6-3, 6-4, 6-5). Similarly if the second die shows $6$ and the first anything but $6$ that can happen another $5$ ways (1-6, 2-6, 3-6, 4-6, 5-6). 
There are $36$ possible rolls of the two dice in total. 
Hence the probability of  exactly one die showing $6$ is
$${10 \over 36} = {5 \over 18}$$

Alternatively: 
If 


*

*$A$ is the event that the first die is $6$ and the second anything

*$B$ is the event that the second die is $6$ and the first anything


then the desired probability is
$$P(\text{One 6})  = P(\text{at least one 6}) - P(\text{two 6s}) $$ $$ \hspace{5 mm} = P(A \cup B) - P(A \cap B)$$ $$ = {11 \over 36} - {1 \over 36} \hspace{15 mm}$$ $$ = {10 \over 36} \hspace{27 mm}$$ 
This is also equal to 
$$P(\text{ first is 6 but not the second }) + P(\text{ second is 6 but not the first })$$
$$ = (P(A) -  P(A \cap B)) + (P(B)-  P(A \cap B))$$
$$ = P(A) + P(B) - 2P(A\cap B)$$
$$= {6 \over 36} + {6 \over 36} - 2 \cdot {1 \over 36}$$
