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1.5.1 Suppose that we roll four fair six-sided dice. (a) What is the conditional probability that the first die shows $2$, conditional on the event that exactly three dice show $2$?


My attempt:

Let A be the event that first die shows 2

Let B be the event that exactly three die shows 2

There is a $\frac{1}{6}$ chance to get 2.

$|S| = 6^4$ total amount of outcomes

How do I go about this?

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You don't need to know anything about dice (including their number of sides) to solve this. You know that exactly three dice show $2$. How many ways are there of choosing $3$ out of $4$ dice? These are all equiprobable. In how many of them does the first die show a $2$?

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    $\begingroup$ I was writing this up mathematically when you beat me to the posting (+1). Nevertheless, I'll post my (i.e., "your") solution to help the OP. $\endgroup$ – David G. Stork Jul 3 '18 at 20:29
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Following @joriki: There is some probability of getting exactly three $2$s, and each of the configurations is equally likely. This means that the non-$2$ can be in one of four positions. In $1/4$ of the cases, then, the first die is the non-$2$ and in $3/4$ of the cases the first die is a $2$. So the ratio is $3$, or the relative probability is $3/4$.

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$B$ has $4\cdot 5=20$ elements, the $4$ to a possible position for the not-$2$, each $5$ to one of the values $1,3,4,5,6$ on this position. We compute $A\cap B$ by computing the complement $B-A$ of it in $B$, $5$ elements. The conditional probability is thus $15/20$.

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