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 
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 
.
![[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]](http://latex.artofproblemsolving.com/f/f/d/ffd6c029970fcec169cae363da55912394de8155.png)
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]](http://latex.artofproblemsolving.com/6/b/c/6bc396866dc2411fccbff425f400cab019e21ab7.png)
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 ![\[\sum\limits_{i=0}^3 {{2+i} \choose 2} \left( \frac{1}{2} \right)^{3+i} = \frac{1}{8} + \frac{3}{16} + \frac{6}{32} + \frac{10}{64} = \boxed{\frac{21}{32}}.\]](http://latex.artofproblemsolving.com/7/7/8/778490c8388b1a251c07f2cf51fc6636171e247e.png)
