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 \cup A_2 \cup \dots \cup A_{10})= 10 \choose 1 | + | <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 \cup A_2 \cup \dots \cup A_{10})</math> |
Revision as of 14:23, 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
$P(A_1 \cup A_2 \cup \dots \cup A_{10})= 10 \choose 1 P(A_1) - 10 \choose 2 P(A_1 \cup A_2)+ \dots - 10 \choose 10 P(A_1 \cup A_2 \cup \dots \cup A_{10})$ (Error compiling LaTeX. Unknown error_msg)