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

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<math>\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</math>
 
<math>\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</math>
  
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(Clarification: the <math>\text{nth}</math> question has <math>n</math> answer choices, where <math>n</math> goes from <math>1</math> to <math>20</math>)
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(Clarification: the <math>n\text{th}</math> question has <math>n</math> answer choices, where <math>n</math> goes from <math>1</math> to <math>20</math>)
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==Solution==
 
==Solution==
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The highest probability occurs when the first <math>8</math> problems are correct. The probability is thus <cmath>\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}</cmath> This, when simplified, gives <cmath>\frac{1}{7!\cdot20}</cmath> Which is in none of the answer choices, so the answer is <math>\boxed{\text{(L) }\text{none of the above}}</math>
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==See Also==
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 +
{{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]]}}

Latest revision as of 22:43, 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

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 $\boxed{\text{(L) }\text{none of the above}}$

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
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