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.

  • 4
    $\begingroup$ There are $4$ equally probable ways to pass the test, namely $444X,44X4,4X44,X444$ which makes the answer... $\endgroup$ – lulu Sep 12 '18 at 23:24
  • $\begingroup$ Note: your question is very confusing. You switch from $4$ to $5$ between the header and the body. Please edit to ask a coherent question. I was responding to part of the question in the body, but you switch things around so much it's hard to know if my comment was relevant at all. $\endgroup$ – lulu Sep 12 '18 at 23:26
  • $\begingroup$ @lulu sorry that was a typo, it should be 5 everywhere. There is only one die which it doesn't start with 5, so the answer is 3/4. I don't even think I was supposed to use the equation here. Is this correct? Thank you $\endgroup$ – K Split X Sep 12 '18 at 23:37
  • 2
    $\begingroup$ No problem. My comment was correct then, except that I ought to have had a $5$ wherever I placed a $4$. Yes, no equation was needed. $\endgroup$ – lulu Sep 12 '18 at 23:40

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 $\frac34$.


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 $P(A \vert B)$ $$p(A \vert B) = \frac{3p}{4p} = \frac{3}{4}$$


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 $5$.

  • $\begingroup$ I'm curious as to how you got the intersection? Could you briefly explain? $\endgroup$ – K Split X Sep 13 '18 at 0:04
  • $\begingroup$ Yes that I can explain. As you know, $P(FE) = P(F)(E|F)$ by the multiplication rule and $P(F)$ is $1/6$. The $P(E|F)$ is the probability of getting two 5's and one non 5, for example 551. There are ${3 \choose 2} ={3 \choose 1}$ arrangements (551, 515, 155). $\endgroup$ – HJ_beginner Sep 13 '18 at 4:42

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