Difference between revisions of "2004 AMC 10B Problems/Problem 11"

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For <math>n>2</math> we already have <math>2\times n = n + n > 2 + n</math>, hence all other cases are good.
 
For <math>n>2</math> we already have <math>2\times n = n + n > 2 + n</math>, hence all other cases are good.
  
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> the sum is smaller, satisfying the condition. Therefore the answer is <math>\frac{48}{64} = \boxed{\frac34}</math>.
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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> cases the sum is smaller, satisfying the condition. Therefore the answer is <math>\frac{48}{64} = \boxed{\frac34}</math>.
  
 
===Solution 2===
 
===Solution 2===

Revision as of 14:07, 19 February 2012

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?

$\mathrm{(A) \ } \frac{1}{2} \qquad \mathrm{(B) \ } \frac{47}{64} \qquad \mathrm{(C) \ } \frac{3}{4} \qquad \mathrm{(D) \ } \frac{55}{64} \qquad \mathrm{(E) \ } \frac{7}{8}$

Solutions

Solution 1

We have $1\times n = n < 1 + n$, hence if at least one of the numbers is $1$, the sum is larger. There $15$ such possibilities.

We have $2\times 2 = 2+2$.

For $n>2$ we already have $2\times n = n + n > 2 + n$, hence all other cases are good.

Out of the $8\times 8$ possible cases, we found that in $15+1=16$ the sum is greater than or equal to the product, hence in $64-16=48$ cases the sum is smaller, satisfying the condition. Therefore the answer is $\frac{48}{64} = \boxed{\frac34}$.

Solution 2

Let the two rolls be $m$, and $n$.

From the restriction: $mn > m + n$

$mn - m - n > 0$

$mn - m - n + 1 > 1$

$(m-1)(n-1) > 1$

Since $m-1$ and $n-1$ are non-negative integers between $1$ and $8$, either $(m-1)(n-1) = 0$, $(m-1)(n-1) = 1$, or $(m-1)(n-1) > 1$

$(m-1)(n-1) = 0$ if and only if $m=1$ or $n=1$.

There are $8$ ordered pairs $(m,n)$ with $m=1$, $8$ ordered pairs with $n=1$, and $1$ ordered pair with $m=1$ and $n=1$. So, there are $8+8-1 = 15$ ordered pairs $(m,n)$ such that $(m-1)(n-1) = 0$.

$(m-1)(n-1) = 1$ if and only if $m-1=1$ and $n-1=1$ or equivalently $m=2$ and $n=2$. This gives $1$ ordered pair $(m,n) = (2,2)$.

So, there are a total of $15+1=16$ ordered pairs $(m,n)$ with $(m-1)(n-1) < 1$.

Since there are a total of $8\cdot8 = 64$ ordered pairs $(m,n)$, there are $64-16 = 48$ ordered pairs $(m,n)$ with $(m-1)(n-1) > 1$.

Thus, the desired probability is $\frac{48}{64} = \frac{3}{4} \Rightarrow C$.

See also

2004 AMC 10B (ProblemsAnswer KeyResources)
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
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