Problem (Combinatorics):
Let $A, B$, and $C$ be randomly chosen (not necessarily distinct) integers between 0 and 4 inclusive. Pat and Chris compute the value of $A+B \cdot C$ by two different methods. Pat follows the proper order of operations, computing $A+(B \cdot C)$. Chris ignores order of operations, choosing instead to compute $(A+B) \cdot C$. Compute the probability that Pat and Chris get the same answer.

Solution:
If Pat and Chris get the same answer, then $A+(B \cdot C)=(A+B) \cdot C$, or $A+B C=A C+B C$, or $A=A C$. This equation is true if $A=0$ or $C=1$; the equation places no restrictions on $B$. There are 25 triples $(A, B, C)$ where $A=0,25$ triples where $C=1$, and 5 triples where $A=0$ and $C=1$. As all triples are equally likely, the answer is $\frac{25+25-5}{5^{3}}=\frac{45}{125}=\frac{\mathbf{9}}{\mathbf{2 5}}$.

Final answer: $\frac{9}{25}$