Tossing two dice with sum equal to 4? Exercise:

Throw two dice. Suppose that eye sum are 4.
  Calculate the resulting conditional probability that
a) the first dice gave a 3 . b ) the second dice gave two or fewer
  eyes. c ) both dice showed an odd number of eyes.

I have no idea about how to "tackle" this task. Do you have any hints? I know that the total possible outcomes of these two dices are 36 and that the conditional probability is
$$P(B|A)=P( joint.of.events.A.and.B)/P(A)$$
but I do not know how to use this.
 A: You have $3$ possible events in your probability-space:
 First Die | Second Die
-----------|------------
     1     |      3
-----------|------------
     2     |      2
-----------|------------
     3     |      1


Out of these $3$ events:


*

*There is $1$ event in which the first die gave 3, hence the probability is $\frac13$

*There are $2$ events in which the second die gave 2 or less, hence the probability is $\frac23$

*There are $2$ events in which both dice showed an odd number, hence the probability is $\frac23$

A: Hint:
Let $X$ denote the sum of eyes by throwing two dice and let $E$
be some event. Then
(*) $P\left(E\text{ and } X=4\right)=P\left(E\mid X=4\right)P\left(X=4\right)$
So finding $P\left(E\text{ and }X=4 \right)$ and $P\left(X=4\right)$ leads to finding
$P\left(E\mid X=4\right)$.
Apply this on the events mentioned in a), b) and c).

Let $D_{1}$
be the score of the first die and let $D_{2}$ be the score of the
second die.
If $i,j\in\left\{ 1,\dots,6\right\} $ then $P\left(D_{1}=i\text{ and }D_{2}=j\right)=P\left(D_{1}=i\right)P\left(D_{2}=j\right)=\frac{1}{6}\frac{1}{6}=\frac{1}{36}$
Now let $E$ be the event that the first die gives a $3$ (as described as case a)). 
Then $P\left(E\text{ and }X=4\right)=P\left(D_{1}=3\text{ and }D_{2}=1\right)=\frac{1}{36}$
Also $$P\left(X=4\right)=P\left(D_{1}=3\text{ and }D_{2}=1\right)+P\left(D_{1}=2\text{ and }D_{2}=2\right)+P\left(D_{1}=1\text{ and }D_{2}=3\right)=\frac{3}{36}$$
Substitution of this in (*) allows you to find $P\left(E\mid X=4\right)=\frac13$
