[Math] the conditional probability that the first die shows 5, conditional on the event that exactly three dice show 5

Question

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.

Answer 1

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 $\frac34$.