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Difference between revisions of "Mock AIME 2 2010 Problems/Problem 3"

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==Problem==
 
==Problem==
 
Five gunmen are shooting each other. At the same moment, each randomly chooses one of the other four to shoot. The probability that there are some two people shooting each other can be expressed in the form <math>\frac{a}{b}</math>, where <math>a, b</math> are relatively prime positive integers. Find <math>a+b</math>.
 
Five gunmen are shooting each other. At the same moment, each randomly chooses one of the other four to shoot. The probability that there are some two people shooting each other can be expressed in the form <math>\frac{a}{b}</math>, where <math>a, b</math> are relatively prime positive integers. Find <math>a+b</math>.
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==Solution 1(PIE)==
 
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Let the five people be <math>A, B, C, D,</math> and <math>E</math>. Let <math>A_1</math> denote the event of <math>A</math> and <math>B</math> shooting each other and <math>A_2</math> through <math>A_{10}</math> similarly. Then, <math>P(\text{Some 2 people are shooting each other})=P(A_1 \cup A_2 \cup \dots \cup A_{10})</math>. Since <math>A, B, C, D,</math> and <math>E</math> are indistinguishable, PIE gives us
 
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<math>P(A_1 \cup A_2 \cup \dots \cup A_{10})= 10 \choose 1 P(A_1) - 10 \choose 2 P(A_1 \cup A_2)+ \dots - 10 \choose 10 P(A_1 \int A_2 \int \dots \int A_{10})</math>
 

Latest revision as of 14:18, 31 December 2023

Problem

Five gunmen are shooting each other. At the same moment, each randomly chooses one of the other four to shoot. The probability that there are some two people shooting each other can be expressed in the form $\frac{a}{b}$, where $a, b$ are relatively prime positive integers. Find $a+b$.

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