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If two six sided die with the numbers $1,0,1,2,3,4$ on each of them are rolled. What is the probability that the product of the numbers on the upmost faces is greater than the sum.

I have an answer key that states the answer to this is $11/36$, but I keep getting $8/36$. What am I missing?

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  • $\begingroup$ Can you add 8 cases you are considering $\endgroup$ – Amar Feb 19 '17 at 17:42
  • $\begingroup$ 2,3 - 2,4 - 3,2 - 3,3 - 3,4 - 4,2 - 4,3 - 4,4 $\endgroup$ – user3753 Feb 19 '17 at 17:46
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I'm also getting the answer $8/36$, the $8$ cases being:

$(4,4),(3,4),(4,3),(2,4),(4,2),(3,3),(2,3),(3,2)$

The other answer I can see to this problem is to consider the cases where the product is greater or $\textbf{equal}$ to the sum. In this scenario, other two cases come into play:

$(2,2),(0,0)$

But in this scenario the answer would be $10/36$. Anyway, I think the right thing to say is that either your answer key is wrong, or the question was misenterpreted and I'm missing one case.

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If 0 on any dice it makes product 0. So not greater than sum.

If 1 on any dice then product with any number is not greater than the sum of 1 and number.

So we left with numbers 2,3 and 4. So these numbers give us 8 cases. You already have. So your answer is correct.

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