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Difference between revisions of "1982 AHSME Problems/Problem 25"

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== Problem ==
 
== Problem ==
−
The adjacent map is part of a city: the small rectangles are rocks, and the paths in between are streets.  
+
The adjacent map is part of a city: the small rectangles are blocks, and the paths in between are streets.  
 
Each morning, a student walks from intersection <math>A</math> to intersection <math>B</math>, always walking along streets shown,  
 
Each morning, a student walks from intersection <math>A</math> to intersection <math>B</math>, always walking along streets shown,  
 
and always going east or south. For variety, at each intersection where he has a choice, he chooses with  
 
and always going east or south. For variety, at each intersection where he has a choice, he chooses with  
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\text{(C)}\frac{4}{7}\qquad
 
\text{(C)}\frac{4}{7}\qquad
 
\text{(D)}\frac{21}{32}\qquad
 
\text{(D)}\frac{21}{32}\qquad
−
\text{(E)}\frac{3}{4} </math>
+
\text{(E)}\frac{3}{4} </math>
  
 
==Solutions==
 
==Solutions==
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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>
 +
 +
==See Also==
 +
{{AHSME box|year=1982|num-b=24|num-a=26}}
 +
 +
{{MAA Notice}}

Latest revision as of 18:05, 11 September 2023

Problem

The adjacent map is part of a city: the small rectangles are blocks, and the paths in between are streets. Each morning, a student walks from intersection $A$ to intersection $B$, 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 $\frac{1}{2}$ whether to go east or south. Find the probability that through any given morning, he goes through $C$.

[asy] defaultpen(linewidth(0.7)+fontsize(8)); size(250); path p=origin--(5,0)--(5,3)--(0,3)--cycle; path q=(5,19)--(6,19)--(6,20)--(5,20)--cycle; int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<6; j=j+1) { draw(shift(6*i, 4*j)*p); }} clip((4,2)--(25,2)--(25,21)--(4,21)--cycle); fill(q^^shift(18,-16)*q^^shift(18,-12)*q, black); label("A", (6,19), SE); label("B", (23,4), NW); label("C", (23,8), NW); draw((26,11.5)--(30,11.5), Arrows(5)); draw((28,9.5)--(28,13.5), Arrows(5)); label("N", (28,13.5), N); label("W", (26,11.5), W); label("E", (30,11.5), E); label("S", (28,9.5), S);[/asy]

$\text{(A)}\frac{11}{32}\qquad \text{(B)}\frac{1}{2}\qquad \text{(C)}\frac{4}{7}\qquad \text{(D)}\frac{21}{32}\qquad \text{(E)}\frac{3}{4}$

Solutions

Solution 1

[asy] defaultpen(linewidth(0.7)+fontsize(8)); size(250); path p=origin--(5,0)--(5,3)--(0,3)--cycle; path q=(5,19)--(6,19)--(6,20)--(5,20)--cycle; int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<6; j=j+1) { draw(shift(6*i, 4*j)*p); }} clip((4,2)--(25,2)--(25,21)--(4,21)--cycle); fill(q^^shift(18,-16)*q^^shift(18,-12)*q, black); label("$A$", (6,19), SE); label("$B$", (23,4), NW); label("$C$", (23,8), NW); label("$C_3$", (17,7), SW); label("$C_2$", (17,11), SW); label("$C_1$", (17,15), SW); label("$C_0$", (17,19), SW); filldraw((17,7)--(18,7)--(18,8)--(17,8)--cycle, black); filldraw((17,11)--(18,11)--(18,12)--(17,12)--cycle, black); filldraw((17,15)--(18,15)--(18,16)--(17,16)--cycle, black); filldraw((17,19)--(18,19)--(18,20)--(17,20)--cycle, black); draw((26,11.5)--(30,11.5), Arrows(5)); draw((28,9.5)--(28,13.5), Arrows(5)); label("N", (28,13.5), N); label("W", (26,11.5), W); label("E", (30,11.5), E); label("S", (28,9.5), S);[/asy]

The probability that the student passes through $C$ is the sum from $i=0$ to $3$ of the probabilities that he enters intersection $C_i$ in the adjoining figure and goes east. The number of paths from $A$ to $C_i$ is ${{2+i} \choose 2}$, because each such path has $2$ eastward block segments and they can occur in any order. The probability of taking any one of these paths to $C_i$ and then going east is $(\tfrac{1}{2})^{3+i}$ because there are $3+i$ intersections along the way (including $A$ and $C_i$) where an independent choice with probability $\tfrac{1}{2}$ is made. So the answer is \[\sum\limits_{i=0}^3 {{2+i} \choose 2} \left( \frac{1}{2} \right)^{3+i} = \frac{1}{8} + \frac{3}{16} + \frac{6}{32} + \frac{10}{64} = \boxed{\frac{21}{32}}.\]

See Also

1982 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Problem 26
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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