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 | + | 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 | ||
Line 31: | Line 31: | ||
\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== | ||
Line 69: | Line 69: | ||
==See Also== | ==See Also== | ||
− | {{AHSME box|year=1982|num-b= | + | {{AHSME box|year=1982|num-b=24|num-a=26}} |
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:05, 11 September 2023
Contents
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 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 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.