Three fair, six-sided dice are tossed. What is the probability that the product of the three rolls is 90?
+633 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.
