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?
 A: 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 
$$P(\text{ first is 6 but not the second }) + P(\text{ second is 6 but not the first })$$
$$ = (P(A) -  P(A \cap B)) + (P(B)-  P(A \cap B))$$
$$ = P(A) + P(B) - 2P(A\cap B)$$
$$= {6 \over 36} + {6 \over 36} - 2 \cdot {1 \over 36}$$
A: Label the dice 1 and 2. You want to find the probability that die 1 rolls a six ($p = \frac 16$) while die 2 rolls anything else ($p = \frac 56$) OR vice versa.
The first case is simply $p = \frac 16 \cdot \frac 56 = \frac {5}{36} $.
The second case is simply the mirror of the first, with the same probability. 
So the final probability is $p = (2)(\frac{5}{36}) = \frac {5}{18}$
A: Lets consider two events: A - the first dice gets 6, B - the second. Then $P(A) = P(B) = 1/6$. And $P(A\cap B) = 1/36$. Then the answer: $p = P(A)+P(B)-2\cdot P(A\cap B) = 1/3-1/18 = 5/18.$
A: Total outcomes: $6^2 = 36$.
Allowed outcomes is the complementary of unallowed outcomes which can be split into "both dice < 6" ( case 1 ) "and" "not both dice = 6" ( case 2 ):
$6^2 - 5^2 - 1^2 = 36-25-1 = 10$. We can write this as a binary number matrix with the indices equal to the dice results:
$\left(\begin{array}{cccccc} 1&1&1&1&1&0\\1&1&1&1&1&0\\1&1&1&1&1&0\\1&1&1&1&1&0\\1&1&1&1&1&0\\0&0&0&0&0&2\end{array}\right)$ we see that the number of which are 0 is 10.
