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Three fair dice, colored red, blue, and yellow, are rolled once. We denote it by R, B, Y the numbers appearing on the upper side of the three dice, respectively.

(a) $P(R=B)=\frac{1}{6}$

(b) $P(R\lt B)=\frac{1}{2}(1-P(R=B))=\frac{5}{12}$

(c) $P(R=B=Y)=\frac{1}{36}$

(d) $P(R\lt B\lt Y)$

I know how to do (a), (b), (c). I think (d) is $$\frac{\left(^{6}_{3}\right)}{216}$$

but this problem need us to solve it with (c) and symmetry. How to do it?

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    $\begingroup$ All of your answers are correct. $\endgroup$ – N. F. Taussig Mar 10 at 16:33
  • $\begingroup$ @N.F.Taussig No. It doesn't solve it with symmetry. Also thank you $\endgroup$ – Maggie Mar 10 at 17:01
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You have to find $P(R\lt B\lt Y)$

All possible rolls can be divided into cases

$P(R \lt B \lt Y)$ has $3!=6$ permutations.

So you can subtract rolls in which any $2$ or all $3$ dice show the same number from the total arrangements and then divide by $6$

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  • $\begingroup$ thank you! $\frac{1}{6}(1-P(R=B=Y)-P(R=B\neq Y)\times 3)=\frac{1}{6}(1-\frac{1}{36}-\frac{5}{6\times 6}\times 3)=\frac{5}{54}$. That's correct. $\endgroup$ – Maggie Mar 10 at 16:58
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The way you solve d) is correct.

It can also be solved with inclusion/exclusion and symmetry as:$$\frac1{3!}\left(1-P(R=B\text{ or } R=C\text{ or } C= B)\right)=$$$$\frac16(1-3P(R=B)+3P(R=B=C)-P(R=B=C))=$$$$\frac16(1-3P(R=B)+2P(R=B=C))$$

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