Difference between revisions of "1982 AHSME Problems/Problem 25"
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The probability that the student passes through <math>C</math> is the sum from <math>i=0</math> to <math>3</math> of the probabilities that he enters intersection <math>C_i</math> in the adjoining figure and goes east. The number of paths from <math>A</math> to <math>C_i</math> is <math>{{2+i} \choose 2}</math>, because each such path has <math>2</math> eastward block segments and they can occur in any order. The probability of taking any one of these paths to <math>C_i</math> and then going east is <math>(\tfrac{1}{2})^{3+i}</math> because there are <math>3+i</math> intersections along the way (including <math>A</math> and <math>C_i</math>) where an independent choice with probability <math>\tfrac{1}{2}</math> is made. So the answer is <cmath>\sum\limits_{i=0}^3 {{2+i} \choose 2} \left( \frac{1}{2} \right)^{3+i} = \frac{1}{8} + \frac{3}{16} + \frac{6}{32} + \frac{10}{64} = \boxed{\frac{21}{32}}.</cmath> | The probability that the student passes through <math>C</math> is the sum from <math>i=0</math> to <math>3</math> of the probabilities that he enters intersection <math>C_i</math> in the adjoining figure and goes east. The number of paths from <math>A</math> to <math>C_i</math> is <math>{{2+i} \choose 2}</math>, because each such path has <math>2</math> eastward block segments and they can occur in any order. The probability of taking any one of these paths to <math>C_i</math> and then going east is <math>(\tfrac{1}{2})^{3+i}</math> because there are <math>3+i</math> intersections along the way (including <math>A</math> and <math>C_i</math>) where an independent choice with probability <math>\tfrac{1}{2}</math> is made. So the answer is <cmath>\sum\limits_{i=0}^3 {{2+i} \choose 2} \left( \frac{1}{2} \right)^{3+i} = \frac{1}{8} + \frac{3}{16} + \frac{6}{32} + \frac{10}{64} = \boxed{\frac{21}{32}}.</cmath> | ||
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+ | ==See Also== | ||
+ | {{AHSME box|year=1982|num-b=1|num-a=3}} | ||
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+ | {{MAA Notice}} |
Revision as of 02:05, 22 December 2020
Contents
Problem
The adjacent map is part of a city: the small rectangles are rocks, and the paths in between are streets. Each morning, a student walks from intersection to intersection , always walking along streets shown, and always going east or south. For variety, at each intersection where he has a choice, he chooses with probability whether to go east or south. Find the probability that through any given morning, he goes through .
Solutions
Solution 1
The probability that the student passes through is the sum from to of the probabilities that he enters intersection in the adjoining figure and goes east. The number of paths from to is , because each such path has eastward block segments and they can occur in any order. The probability of taking any one of these paths to and then going east is because there are intersections along the way (including and ) where an independent choice with probability is made. So the answer is
See Also
1982 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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 | ||
All AHSME Problems and Solutions |
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