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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?

$P(first die shows 2) = \frac{1}{6}$

$P( exactly 3 dice show 2) = 1 - P( exactly 3 dice don't show 2)$

$P(first die shows 2|exactly 3 dice shows 2) = \frac{P(first die shows 2 \cap exactly 3 dice shows 2)}{P(eactly 3 dice shows 2)}$

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  • $\begingroup$ The probability that exactly three dice show a two can be found directly using a binomial distribution. Further, the intersection of the events can also be found directly using a binomial distribution noting that if the first die shows a 2 and there are exactly three 2's total among the four dice, that implies that among the second,third, and fourth rolls exactly two of those show 2's. $\endgroup$ – JMoravitz May 15 '18 at 23:28
  • $\begingroup$ Alternatively, you could skip all of the rigorous calculations and compare your problem to the following 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? Convince yourself why the answer to this alternate question is the same as the answer to your original question. $\endgroup$ – JMoravitz May 15 '18 at 23:30
  • $\begingroup$ So the probability of the second, third, and fourth dice showing a two would be the prob of all 4 dice showing a two subtracted by the first one showing a 2? $\endgroup$ – dg123 May 15 '18 at 23:31
  • $\begingroup$ no. No subtraction is needed here at all. $\endgroup$ – JMoravitz May 15 '18 at 23:31
  • $\begingroup$ And the answer to the alternate question is 3 out of 4, but how does that help with the original question? $\endgroup$ – dg123 May 15 '18 at 23:32
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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.

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  • $\begingroup$ Where did 4C3 come from and why are we multiplying by 5 over 6? $\endgroup$ – dg123 May 15 '18 at 23:54
  • $\begingroup$ @dg123 en.wikipedia.org/wiki/Binomial_distribution $\endgroup$ – JMoravitz May 15 '18 at 23:55
  • $\begingroup$ @dg123 $\tbinom 43 \tfrac 5{6^4}$ is the probability that three from four die will be a six and the fourth not. $~~\tfrac 5{6^4}$ is the probability for a particular arrangement that shows the required count of faces, and $\tbinom 43$ counts the ways those dice may be arranged. $\endgroup$ – Graham Kemp May 16 '18 at 0:15
  • $\begingroup$ so 5 over 6 to the 4 is the probability for which the twos can occur in all four dice? $\endgroup$ – dg123 May 16 '18 at 0:29
  • $\begingroup$ @dg123 $\frac{5}{6^4}=\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{5}{6}$ is the probability that the first die is a two, the second die is a two, the third die is a two, and the fourth die is not a two. It is also equal to $\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{5}{6}\cdot\frac{1}{6}$, the probability that the first die is a two, the second die is a two, the third die is not a two, and the fourth die is a two, etc... $\endgroup$ – JMoravitz May 16 '18 at 0:54
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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 ...

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

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