## Problem of the Day #144: Arjun’s Problems
*August 10, 2011*

*Posted by Arjun in : potd , trackback*

Since Albert has been bugging him to write a problem, Arjun decides to pick one out of a set of hats. However, Aziz (also needing a problem) decides to pick one out the same set of hats. There are three hats: the first has $5$ acceptable problems, the second has $6$, and the third has $8$. If Arjun and Aziz each pick 3 problems (each problem has equal chance of being chosen) from the hats (with replacement), the probability that none of Arjun’s problems were from the same hat as any of Aziz’s problems can be represented as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

## Comments»

“with replacement” meaning if Arjun chooses the same problem 3 times then it’s fine?

Yes.

It’s also possible that Aziz and Arjun both choose the same problem 3 times.