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As the question states, I need to find the probability of rolling 3 dice numbers which product is a multiple of 6. what I have tried:

$\omega={(6,k,k), (3,2,k), (3,4,k))}$

$P(A) = \frac{3\times(1^1\times6^2) + 6\times(1^1\times1^1\times6^1) + 6\times(1^1\times1^1\times6^1)}{6^3}=\frac{180}{216}=\frac{5}{6}$

I know writing $1^1$ seems silly, but just trying to show what I'm doing

Thanks for helping and sorry if it is a bad question

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    $\begingroup$ This looks wrong at a glance as it looks like you are overcounting. For instance, where does the outcome where all three dice show a $6$ get counted and how many times did you count it? It should only have been counted once, but it looks like you counted it three times. What about the outcome where you rolled a $2,3,6$? in some order? Was it counted in the first category? or in the second? $\endgroup$
    – JMoravitz
    Jun 14 at 14:41
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    $\begingroup$ You have overcounted since there is an overlap between the cases you describe. E.g. (6,2,3) is in two of the cases so is counted twice. $\endgroup$ Jun 14 at 14:42
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    $\begingroup$ I think it would be easier to count the outcomes where the product is not even, count the outcomes where the product is not a multiple of three, and count the outcomes where the product is neither even nor a multiple of three, and use that to proceed to a solution. $\endgroup$
    – JMoravitz
    Jun 14 at 14:43
  • $\begingroup$ @JMoravitz what if i use the same way i did but substract the cases where i overcounted? $\endgroup$ Jun 14 at 14:47
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    $\begingroup$ That seems tedious. The question is crafted in such a way as to specifically suggest using inclusion-exclusion and basic facts about divisibility to approach. If you want to explore other additional approaches after the fact, go ahead and revisit the problem, but I strongly encourage you to use the approach I suggested in my second comment first. $\endgroup$
    – JMoravitz
    Jun 14 at 14:48
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Let the dice be thrown in sequence or otherwise be distinguishable some other way. There are $6^3 = 216$ total different equally likely outcomes.

Approach by trying to count how many of these outcomes were "bad" and should be removed because they either are not multiples of $2$, are not multiples of $3$, and keeping in mind that they might have been both.

Of these, $3^3=27$ are "bad" because their product of the results is not even (and thus not a multiple of six), seen by noting a product is odd iff all terms in the product is odd, there being three odd numbers.

$~$

Similarly, $4^3=64$ are "bad" because their product of results is not a multiple of three (and thus not a multiple of six), seen by noting a product is not a multiple of three iff all terms in the product are not multiples of three, there being four possible non-multiple of three numbers to choose from here.

$~$

Finally, some of these "bad" outcomes we counted twice as they were simultaneously not even and not multiples of three. These would be the outcomes consisting only of $1$'s and $5$'s, there being $2^3=8$ of which. We keep this in mind to correct our count as we proceed with inclusion-exclusion.

The probability then is:

$$\frac{6^3-3^3-4^3+2^3}{6^3} = \frac{216 - 27 - 64 + 8}{216} = \frac{133}{216}$$

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Let $a_2,b_2,c_2$ ($a_3,b_3,c_3$) denote the exponents of $2$ ($3$) in the prime factorization of the number shown on the three dice. Then the probability you want is

$$P(a_2+b_2+c_2>0, a_3+b_3+c_3>0) = 1-P(a_2+b_2+c_2=0\ or \ a_3+b_3+c_3=0)$$

$$ = 1-P(a_2+b_2+c_2=0)-P(a_3+b_3+c_3=0)+P(a_2+b_2+c_2=0, a_3+b_3+c_3=0)$$

$$ (by\ independence)= 1-P(a_2=0)^3-P(a_3=0)^3+P(a_2=0,a_3=0)^3$$

$$ = 1-P(1,3,\ or \ 5)^3-P(1,2,4, \ or \ 5)^3+P(1 \ or \ 5)^3$$

$$ = 1 - (\frac{3}{6})^3 - (\frac{4}{6})^3+(\frac{2}{6})^3 = \frac{133}{216}.$$

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    $\begingroup$ very elegant approach ! $\endgroup$
    – Bulbasaur
    Jun 14 at 15:32

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