Difference between revisions of "2006 iTest Problems/Problem 12"

m (Problem)
m
Line 10: Line 10:
  
 
==Solution==
 
==Solution==
 +
 +
==See Also==
 +
 +
{{iTest box|year=2006|num-b=11|num-a=13|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Problems/Problem U2|U2]] '''•''' [[2006 iTest Problems/Problem U3|U3]] '''•''' [[2006 iTest Problems/Problem U4|U4]] '''•''' [[2006 iTest Problems/Problem U5|U5]] '''•''' [[2006 iTest Problems/Problem U6|U6]] '''•''' [[2006 iTest Problems/Problem U7|U7]] '''•''' [[2006 iTest Problems/Problem U8|U8]] '''•''' [[2006 iTest Problems/Problem U9|U9]] '''•''' [[2006 iTest Problems/Problem U10|U10]]}}

Revision as of 23:34, 3 November 2023

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

See Also

2006 iTest (Problems, Answer Key)
Preceded by:
Problem 11
Followed by:
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 U1 U2 U3 U4 U5 U6 U7 U8 U9 U10