How to apply Conditional Probability? For this question, I'm not sure how to use the Conditional Probability formula. Here is what I have so far. Can anyone please help me out?
Suppose that we roll four fair six-sided dice. What is the conditional probability that the first die shows 2, conditional on the
event that exactly three dice show 2?
$P(first die shows 2) = \frac{1}{6}$
$P( exactly 3 dice show 2) = 1 - P( exactly 3 dice don't show 2)$
$P(first die shows 2|exactly 3 dice shows 2) = \frac{P(first die shows 2 \cap exactly 3 dice shows 2)}{P(eactly 3 dice shows 2)}$
 A: Let $A$ represent the event that the first die shows a $2$.
Let $B$ represent the event that among the four dice, exactly three of them show a $2$ and the remaining die doesn't.
First, let us calculate $Pr(A)$.  This you should know from the definition of a "fair die" and from example.
$Pr(A)=\frac{1}{6}$
Next, let us calculate $Pr(B)$.  For this, we recognize it as a binomial distribution problem where there are $n=4$ independent trials, $k=3$ successes, and $p=\frac{1}{6}$ chance for success on each trial.
$Pr(B)=\binom{4}{3}\left(\dfrac{1}{6}\right)^3\left(\dfrac{5}{6}\right)^1=\dfrac{4\cdot 5}{6^4}$
Now, to calculate $Pr(A\cap B)$, we can expand this using $Pr(A)\cdot Pr(B\mid A)$ as per the definition of conditional probability.  From here, note that $Pr(B\mid A)$ is again another binomial distribution question as our attention is entirely focused on the second, third, and fourth rolls, making it $n=3$ independent trials, $k=2$ successes within those three trials, and $p=\frac{1}{6}$ chance for success.
$Pr(A\cap B)=Pr(A)\cdot Pr(B\mid A)=\dfrac{1}{6}\cdot\binom{3}{2}\left(\dfrac{1}{6}\right)^2\left(\dfrac{5}{6}\right)^1=\dfrac{3\cdot 5}{6^4}$

Now, combining all of this information together:
$Pr(A\mid B)=\dfrac{Pr(A\cap B)}{Pr(B)}=\dfrac{\left(\dfrac{3\cdot 5}{6^4}\right)}{\left(\dfrac{4\cdot 5}{6^4}\right)}=\dfrac{3}{4}$

As mentioned in the comments, you could have skipped all of this troublesome work by comparing this to the problem: "You have an urn with three red balls and one green ball. You then draw one of these balls uniformly at random. What is the probability that it is red?" to which the answer is immediately $\frac{3}{4}$, as expected.
A: Let $F$ be the event that "the first die shows 2", and $E$ be the event that "exactly three from the four dice show 2".   Then what you seek is indeed:$$\mathsf P(F\mid E)=\dfrac{\mathsf P(F\cap E)}{\mathsf P(E)}$$
Since the count for shown 2 will be Binomially distributed among however many were thrown (with success rate $1/6$), we can see that:
$${\mathsf P(F)=\dfrac 16\\\mathsf P(E)=\dbinom 43 \dfrac{5}{6^4}}$$
Now, $F\cap E$ is the event that "the first die shows 2 and exactly two from the other three dice show 2".$$\mathsf P(F\cap E)=\dfrac 16\dbinom 32\dfrac 5{6^3}$$
The rest is just substitution and algebra.

Alternatively $\mathsf P(F\mid E)$ is the probability that the first die is among the exactly three from four dice that shows 2, so ...

 $\mathsf P(F\mid E)=3/4$.

