A blue die and a red die are rolled - events A-F shows the following: A: The red die shows even number (27/36...I think?)
B: The blue die shows odd number (also 27/36, I believe)
C: The sum of the face numbers is 6 (5/36)
D: The sum of the face numbers is 9 (4/36)
E: At least one die shows 2 (Not sure, but I think it's 11/36)
F: The difference of 2 faces is one. (Not sure, but I think it's 6/36)
I need to find (P=probability): 
(a) P(A and B)
-For this one, I'm not sure if I got the probabilities (above) right, so I got 18/36. 
(b) P(F and D) 
-I got 1/36
(c) P(E or F)
-I got 14/36, not sure if it's correct
(d) P(F given E)
(e) P(A given E)
Now, for the conditional probabilities (d and e), I have no idea how to do. What is your approach to getting the answers to the conditional probabilities? Also, can someone make sure the answers I got were correct, and if not, explain why it isn't correct for future reference? 
Thank you!
 A: A : The red die shows even number.
This event concerns only the red die and is indeed independant of the blue die hence :
$$P(A) = {3\over 6} = {1\over 2}$$
B : The blue die shows odd number.
For the same reasons as event $A$ :
$$P(B) = {1\over 2}$$
C : The sum of the face numbers is $6$.
Here we can solve this by enumerating the number of relevant events over the total number of possible events.
Here are the $(blue, red)$ events that satisfiy $blue + red = 6$ :
$$(1,5) , (2,4), (3,3), (4,2),(5,1)$$
We can see there are $5$ different cases that satisfy the condition and we know that the total number of combinations is $36$ so we can conclude :
$$P(C) = {5\over 36}$$
D : The sum of the face numbers is 9
Same as $C$, answer is :
$$P(D) = {4\over 36}$$
E : At least one die shows $2$
Let's enumerate the possibilities :
$$(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(1,2),(3,2),(4,2),(5,2),(6,2)$$
So $$P(E)={11\over 36}$$
F : The difference of two face is one.
Again we enumerate favorable rolls :
$$(1,2),(2,1)(2,3),(3,2)(3,4),(4,3)(4,5),(5,4),(5,6),(6,5)$$
So :
$$ P(F) = {10\over 36} $$
(a) $P(A\cap B)$
Since we saw that $A$ and $B$ are independant events, we have :
$$P(A\cap B) = P(A)*P(B) = {1\over 4}$$
(b) $P(F\cap D)$
Let's look at the couples that satisfy $D$ :
$$(3,6),(4,5),(5,4),(6,3)$$
We can see that only $(4,5)$ and $(5,4)$ satisfy $F$ so :
$$P(F\cap D) = {2\over 36} = {1\over 18}$$
(c) $P(E\cup F)$
For this we can apply the formula :
$$P(E\cup F) = P(E)+P(F)-P(E\cap F)$$
We have $P(E) = {11\over 36}$, $P(F) = {10\over 36}$ and we can see that $P(E\cap F) = {4\over 36}$ because $E\cap F$ is satisfied by the four events : $(2,1),(2,3),(1,2),(3,2)$
So we have :
$$P(E\cup F) = {11+10-4\over 36}={17\over 36}$$
(d) $P(F|E)$
We can use the following formula :
$$P(F|E) = {P(F\cap E)\over P(E)} = {{4\over 36}\over {11\over 36} }= {4\over 36}*{36\over 11} = {4\over 11}$$
(e) $P(A|E)$
Same thing : $$P(A|E) = {P(A\cap E)\over P(E)}$$
We can again enumerate the cases that satisfy $A\cap E$ :
$$(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(4,2),(6,2)$$
So $P(A\cap E) = {8\over 36}$
And :$$P(A|E) = {{8\over 36}\over {11\over 36}} = {8\over 36}*{36\over 11} = {8\over 11}$$
A: $P(A)=P(B)=18/36=1/2$ because there are $36$ possibilities and $18$ of those have red showing even and $18$ have blue showing odd.
$P(C)$, $P(D)$, and $P(E)$ are correct.
$P(F)=10/36=5/18$ because $(1,2),(2,3),(3,4),(4,5),(5,6),(6,5),(5,4),(4,3),(3,2),(2,1)$ are all of the possibilities of having a difference of $1$ between the two dice.
(a) $P(A\cap B)=9/36$ because the red die can only be $2,3,4$ and the blue die can only be $1,3,5$, so there are $3^2=9$ possibilites of having both $A$ and $B$.
(b) $P(F\cap D)=2/36=1/18$ because, as seen in the enumeration above when calculating $P(F)$, only $(4,5)$ and $(5,4)$ sum to $9$ and have a difference of $1$.
(c) $P(E\cup F) = P(E)+P(F)-P(E\cap F)$. Now $P(E\cap F)=4/36=1/9$ by again looking at the enumeration when calculating $P(F)$.
For conditional probabilities, you want to use the definition
$$ P(X \mid Y) = \frac{P(X\cap Y)}{P(Y)}.$$
(d) $P(F \mid E) = \frac{P(F\cap E)}{P(E)}$. We have already found both $P(E\cap F)$ and $P(E)$, so this one is easy.
(e) $P(A \mid E) = \frac{P(A\cap E)}{P(E)}$. We know $P(E)$, so we need only compute $P(A\cap E)$. I'll let you do this.
