What is the conditional probability that the first die shows 5, conditional on the event that exactly three dice show 5? Suppose that we roll four fair six-sided dice.
What is the conditional probability that the first die shows 5, conditional on the
event that exactly three dice show 5?
Let $A=\{\text{first dice shows 5}\}$
Let $B=\{\text{3 dice shows 5}\}$
We want $P(A|B)=P(A\cap B)/P(B)$
I know that the size of the sample space $S=6^4$ , but I don't know how to compute $P(A \cap B)$
The $P(B)=(1/6)^3$, since for each dice its $1/6$ chance of showing $5$.
I am stuck on the intersection part.
 A: You're overthinking this. The probability that the first die is the one not showing a $5$ is $\frac14$ by symmetry. Hence the first die is showing a $5$ with probability $\frac34$.
A: Computing $P(B)$
$P(B) $ is the probability that $3$ dice shows $5$. You've got the following possible outcomes (as @lulu mentions)


*

*$$555-$$

*$$55-5$$

*$$5-55$$

*$$-555$$
The probability of any of the above events is denoted by $p = \frac{1}{6^3}$
Therefore, $$P(B) = 4p$$


Computing $P(A \cap B)$
$P(A \cap B)$ is the probability that $3$ dice will show $5$ and first dice shows $5$. The possible outcomes are the above except the the last one, so 
$$p(A \cap B) = 3p$$
Computing $P(A \vert B)$
$$p(A \vert B) = \frac{3p}{4p} = \frac{3}{4}$$ 
A: Let $F$ be the event the first die is $5$.
Let $E$ be the event that three out of four dice show $5$.
If you want to solve it using conditioning...
$$P(F \mid E) = \frac{P(FE)}{P(E)} = \frac{P(F)P(E|F)}{P(E)}= \frac{\left(\frac{1}{6}\right) {3 \choose 2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)}{{4 \choose 1} \left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)} = \frac{3}{4}$$
Word of caution I am a beginner so let me know if you disagree. Basically using the pmf of a Binomial RV. For example exactly $3$ out of $4$ dice are outcome $5$.
