Conditional probability rolling two dice Two fair dice are rolled. What is the conditional
probability that at least one lands on 6 given that
the dice land on different numbers?
I already know the answer, but am having some trouble understanding it. 
If E is the event that at least one dice lands on 6 and F is the event that the dice land on different numbers, I need to calculate P(EF)/P(F)
According to the answer P(EF) = 2*(1/6)*5/6. I don't understand how this was calculated. Is there a specific formula for calculating P(EF)? I know P(EF) = P(E)+P(F) if the two events are mutually exclusive. 
P(F) = 30/36 which makes sense because there are 36 possible outcomes, of which 30 are favorable. 
Mainly just confused about calculating P(EF).
 A: (SIde note: The events are neither mutually exclusive nor independent.)
$P(E\cap F)$ is the probability that one die is a six and the other die is not.  That's $\frac 1 6\frac 5 6+\frac 5 6\frac 1 6$ by adding the probability that the first die is a six and the other not, to the probability that the first die is not a six and the other is. (NB: Those events are mutually exclusive partitions of $E\cap F$.)
$$\mathsf P(E\cap F) = 2 \cdot \frac 1 6 \cdot \frac 5 6\\ = \frac{10}{36}$$
Then we just use conditional probability as you noted.
$$\mathsf P(E\mid F) = \frac{\mathsf P(E\cap F)}{\mathsf P(F)}\\ = \frac{10/36}{30/36} \\ = \frac {1}{3}$$
A: $$P(E|F)=\frac{P(E\cap F)}{P(F)}.$$
Here $P(E\cap F)$= number of pairs $(x,y)$, where $x$ or $y$ are $6$ but not both $6$. This probability is just $10/36$. Note that $2\frac 16 (56)>1$.
A: It might be useful to use a symmetry argument to count the size of EF.  EF has 30 pairs of numbers.  That's a total of 60 numbers.  Since each of the six choices shows up exactly the same number of times (there's the symmetry argument), 1/6*60 = 10 of those 60 numbers is the number '6'.  And each of those 10 is in a different pair, based on how F is defined.  So that's how you get 10/30.
