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

## 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 $A$ 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 $$\sum\limits_{i=0}^3 {{2+i} \choose 2} \left( \frac{1}{2} \right)^{3+i} = \frac{1}{8} + \frac{3}{16} + \frac{6}{32} + \frac{10}{64} = \boxed{\frac{21}{32}}.$$