We have 3 dice labeled R,B,G for the color. We say a Die is maximal if it's larger or equal to the others.

Question

We have 3 dice labeled R,B,G for the color. We say a Die is maximal if it's larger or equal to the others.

If any of my thinking is wrong please point it out

a)What is the probability that the maximum is 4?

$P(R=4, B\le4, G\le4)$ and since the events are independent it's equal to $P(R=4)P(B\le4)P(G\le4)=(1/6)(4/6)(4/6)$. Then we have to do this for the case of R,G or B being true for all 3. So that's
$P(R=4, B\le4, G\le4)\cup P(B=4, R\le4, G\le4)\cup P(G=4, B\le4, R\le4)$
=$P(R=4, B\le4, G\le4) + P(R=4, R\le4, G\le4) + P(G=4, B\le4, R\le4)$ -$P(R=4, B=4, G\le4) + P(R=4, R\le4, G=4) + P(G\le4, B=4, R=4)$ +$P(R=4, B=4, G=4)$
By inclusion-exclusion rule.

b)What is the probability that R is the maximal with R=4?

Just calculate by doing all the events of R=4 and B and G being $\le 4$. So therefore.
$P(R max|R=4)=P(R max, R=4)/P(R=4)$
=$(4^2/6^3)/(6^2/6^3)$
$4^2/6^2$

c)What is the probability that R=4 conditioned on event that R is maximal?

P(R=4|R is maximal)=P(R=4, R is maximal)/P(R is Maximal)
$(4^2/6^3)/\sum^6_{i=1}i^2/6^3$

Answer 1

There are a couple of problems with a).

First, the number of ways to roll at most $4$ with a given die is $4$, not $\binom64$. So your calculation of $\binom64\binom64\big/6^3$ for the probability that (say) the red die rolls $4$ and the others roll $\leq 4$ should be ${4^2}\big/{6^3}$.

Secondly, there is a more subtle problem. You can't just multiply this by $3$ to get the total probability, because you have counted some situations more than once. For example, the situation $R=4,G=4,B=2$ is included in the probability for $R=4, G\leq 4, B\leq 4$ but also in the probability for $G=4, R\leq 4, B\leq 4$. To deal with this you need to calculate the following three situations separately:

• one die shows $4$, the others are $\leq 3$;
• two dice show $4$, the other one is $\leq 3$;
• all three dice show $4$.

Note that these cases cover all possibilities you are interested in, and have no overlap, so you can then just add up the three probabilities.

There are similar problems with the other parts.

Answer 2

This answer does not deal with flaws in your answer (for that see the other answers) but gives you some hints how to do it correctly.

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Let $R,G,B$ be random variables such that e.g. $R$ denotes the result for the die labeled with R.

Then $R,G,B$ are iid and uniformly distributed on $\left\{ 1,2,3,4,5,6\right\} $.

Let $M:=\max\left\{ R,G,B\right\} $.

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Hints on a):

• $P\left(M=4\right)=P\left(M\leq4\right)-P\left(M\leq3\right)$.
• $\{M\leq m\}=\{R\leq m, G\leq m, B\leq m\}$

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Hint on b):

• $P\left(R=M=4\right)=P\left(R=4,G\leq4,B\leq4\right)$

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Hints on c):

• $P\left(R=4\mid M=R\right)=P\left(R=4=M\right)/P\left(M=R\right)$
• $P\left(M=R\right)=P\left(G\leq R,B\leq R\right)=\sum_{r=1}^{6}P\left(G\leq r,B\leq r\mid R=r\right)P\left(R=r\right)$

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Especially exploit the independence of $R,G,B$.