# 2006 iTest Problems/Problem 12

## Problem

What is the highest possible probability of getting $12$ of these $20$ multiple choice questions correct, given that you don't know how to work any of them and are forced to blindly guess on each one?

$\text{(A) }\frac{1}{6!}\qquad \text{(B) }\frac{1}{7!}\qquad \text{(C) }\frac{1}{8!}\qquad \text{(D) }\frac{1}{9!}\qquad \text{(E) }\frac{1}{10!}\qquad \text{(F) }\frac{1}{11!}\qquad\\ \\ \text{(G) }\frac{1}{12!}\qquad \text{(H) }\frac{2}{8!}\qquad \text{(I) }\frac{2}{10!}\qquad \text{(J) }\frac{2}{12!}\qquad \text{(K) }\frac{1}{20!}\qquad \text{(L) }\text{none of the above}\qquad$

(Clarification: the $n\text{th}$ question has $n$ answer choices, where $n$ goes from $1$ to $20$)

## Solution

The highest probability occurs when the first $8$ problems are correct. The probability is thus $$\frac{1}{1}\cdot\frac{1}{2}\cdot\frac{1}{3}\dots\frac{1}{8}\cdot\frac{8}{9}\cdot\frac{9}{10}\dots\frac{18}{19}\cdot\frac{19}{20}$$ This, when simplified, gives $$\frac{1}{7!\cdot20}$$ Which is in none of the answer choices, so the answer is $\boxed{\text{(L) }\text{none of the above}}$