Difference between revisions of "Mock AIME 2 2010 Problems/Problem 3"
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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 | 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 | ||
| − | <math>P(A_1\cupA_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> | + | <math>P(A_1\cupA_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})=10 \cdot P(A_1)</math> |
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| + | Since the probability of <math>A</math> and <math>B</math> shooting each other is just <math>\frac{1}{16}</math>, the desired probability is <math>\frac{5}{8}</math> and the desired sum is <math>\boxed{013}</math>. | ||
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| + | Note how this don't match the answer key. I made a mistake somewhere but don't know where. | ||
Revision as of 14:16, 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 
, where 
are relatively prime positive integers. Find 
.
Solution 1(PIE)
Let the five people be 
and 
. Let 
denote the event of 
and 
shooting each other and 
through 
similarly. Then, 
. Since and are indistinguishable, PIE gives us
$P(A_1\cupA_2\cup\dots\cup A_{10})=10\choose 1 P(A_1) - 10 \choose 2 P(A_1 \cup A_2)+ \dots - 10 \choose 10 P(A_1 \int A_2 \int \dots \int A_{10})=10 \cdot P(A_1)$ (Error compiling LaTeX. Unknown error_msg)
Since the probability of and shooting each other is just 
, the desired probability is 
and the desired sum is 
.
Note how this don't match the answer key. I made a mistake somewhere but don't know where.
