Difference between revisions of "2004 AMC 10B Problems/Problem 11"
(New page: == Problem == Two eight-sided dice each have faces numbered 1 through 8. When the dice are rolled, each face has an equal probability of appearing on the top. What is the probability that...) |
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Out of the <math>8\times 8</math> possible cases, we found that in <math>15+1=16</math> the sum is greater than or equal to the product, hence in <math>64-16=48</math> it is smaller. Therefore the answer is <math>\frac{48}{64} = \boxed{\frac34}</math>. | Out of the <math>8\times 8</math> possible cases, we found that in <math>15+1=16</math> the sum is greater than or equal to the product, hence in <math>64-16=48</math> it is smaller. Therefore the answer is <math>\frac{48}{64} = \boxed{\frac34}</math>. | ||
+ | |||
+ | ==Solution2== | ||
+ | Let the two rolls be <math>m</math>, and <math>n</math>. | ||
+ | |||
+ | From the restriction: | ||
+ | <math>mn > m + n</math> | ||
+ | |||
+ | <math>mn - m - n > 0</math> | ||
+ | |||
+ | <math>mn - m - n + 1 > 1</math> | ||
+ | |||
+ | <math>(m-1)(n-1) > 1</math> | ||
+ | |||
+ | Since <math>m-1</math> and <math>n-1</math> are non-negative integers between <math>1</math> and <math>8</math>, either <math>(m-1)(n-1) = 0</math>, <math>(m-1)(n-1) = 1</math>, or <math>(m-1)(n-1) > 1</math> | ||
+ | |||
+ | <math>(m-1)(n-1) = 0</math> if and only if <math>m=1</math> or <math>n=1</math>. | ||
+ | |||
+ | There are <math>8</math> ordered pairs <math>(m,n)</math> with <math>m=1</math>, <math>8</math> ordered pairs with <math>n=1</math>, and <math>1</math> ordered pair with <math>m=1</math> and <math>n=1</math>. So, there are <math>8+8-1 = 15</math> ordered pairs <math>(m,n)</math> such that <math>(m-1)(n-1) = 0</math>. | ||
+ | |||
+ | <math>(m-1)(n-1) = 1</math> if and only if <math>m-1=1</math> and <math>n-1=1</math> or equivalently <math>m=2</math> and <math>n=2</math>. This gives <math>1</math> ordered pair <math>(m,n) = (2,2)</math>. | ||
+ | |||
+ | So, there are a total of <math>15+1=16</math> ordered pairs <math>(m,n)</math> with <math>(m-1)(n-1) < 1</math>. | ||
+ | |||
+ | Since there are a total of <math>8\cdot8 = 64</math> ordered pairs <math>(m,n)</math>, there are <math>64-16 = 48</math> ordered pairs <math>(m,n)</math> with <math>(m-1)(n-1) > 1</math>. | ||
+ | |||
+ | Thus, the desired probability is <math>\frac{48}{64} = \frac{3}{4} \Rightarrow C</math>. | ||
== See also == | == See also == | ||
{{AMC10 box|year=2004|ab=B|num-b=10|num-a=12}} | {{AMC10 box|year=2004|ab=B|num-b=10|num-a=12}} |
Revision as of 14:15, 7 February 2009
Contents
[hide]Problem
Two eight-sided dice each have faces numbered 1 through 8. 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
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 found that in the sum is greater than or equal to the product, hence in it is smaller. Therefore the answer is .
Solution2
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 |