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# I need help!

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Three fair, six-sided dice are tossed. What is the probability that the product of the three rolls is 90?

Sep 2, 2021

#1
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Let's first find the ways that we could get products of 90.

One of the obvious first choices would be 6, as it's a factor of 90 and reduces it the most greatly. Let's start with it.

90/6 = 20

Now we can factor 20 into 5 and 4, so we now know that the only valid combination is 6-5-4.

We need the total number of permutations where we get a 6-5-4. In this case, our set is {6 5 4} and we are picking 3. This could be written as "3 pick 3", which happens to be equal to 3!. (3 factorial, not just an excited 3 :) )

Now let's put this over the total number of perumations of dice:

$$\frac{3!}{6^3} = \frac{6}{216} = \frac{1}{36}$$

Thus, the chances of the dice having a product of 90 is 1/36.

Sep 2, 2021
#2
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Good technique, but there's a little hiccup:  90 / 6 = 15 not 20.  You get the same answer, though.

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Guest Sep 2, 2021