Difference between revisions of "2004 AMC 10B Problems/Problem 11"
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<math>\mathrm{(A)\ }\frac{1}{2}\qquad\mathrm{(B)\ }\frac{47}{64}\qquad\mathrm{(C)\ }\frac{3}{4}\qquad\mathrm{(D)\ }\frac{55}{64}\qquad\mathrm{(E)\ }\frac{7}{8}</math> | <math>\mathrm{(A)\ }\frac{1}{2}\qquad\mathrm{(B)\ }\frac{47}{64}\qquad\mathrm{(C)\ }\frac{3}{4}\qquad\mathrm{(D)\ }\frac{55}{64}\qquad\mathrm{(E)\ }\frac{7}{8}</math> | ||
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== Solution 1 == | == Solution 1 == | ||
Revision as of 00:36, 24 July 2014
Contents
Problem
Two eight-sided dice each have faces numbered through
. When the dice are rolled, each face has an equal probability of appearing on the top. What is the probability that the product of the two top numbers is greater than their sum?
Solution 1
We have , hence if at least one of the numbers is
, the sum is larger. There
such possibilities.
We have .
For we already have
, hence all other cases are good.
Out of the possible cases, we find that in
the sum is greater than or equal to the product, hence in
cases the sum is smaller, satisfying the condition. Therefore the answer is
.
Solution 2
Let the two rolls be , and
.
From the restriction:
Since and
are non-negative integers between
and
, either
,
, or
if and only if
or
.
There are ordered pairs
with
,
ordered pairs with
, and
ordered pair with
and
. So, there are
ordered pairs
such that
.
if and only if
and
or equivalently
and
. This gives
ordered pair
.
So, there are a total of ordered pairs
with
.
Since there are a total of ordered pairs
, there are
ordered pairs
with
.
Thus, the desired probability is .
See also
2004 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.