2008 AMC 10B Problems/Problem 20

Problem

The faces of a cubical die are marked with the numbers $1$ , $2$ , $2$ , $3$ , $3$ , and $4$ . The faces of another die are marked with the numbers $1$ , $3$ , $4$ , $5$ , $6$ , and $8$ . What is the probability that the sum of the top two numbers will be $5$ , $7$ , or $9$ ?

$\mathrm{(A)}\ 5/18\qquad\mathrm{(B)}\ 7/18\qquad\mathrm{(C)}\ 11/18\qquad\mathrm{(D)}\ 3/4\qquad\mathrm{(E)}\ 8/9$

Solution 1

The easiest way is to write a table of all $36$ possible outcomes, do the sums, and count good outcomes.

     1  3  4  5  6  8
   ------------------
1 |  2  4  5  6  7  9
2 |  3  5  6  7  8 10
2 |  3  5  6  7  8 10
3 |  4  6  7  8  9 11
3 |  4  6  7  8  9 11
4 |  5  7  8  9 10 12

We see that out of $36$ possible outcomes, $4$ give the sum of $5$ , $6$ the sum of $7$ , and $4$ the sum of $9$ , hence the resulting probability is $\frac{4+6+4}{36}=\frac{14}{36}=\boxed{\frac{7}{18}}$ .

Solution 2

Each die is equally likely to roll odd or even, so the probability of an odd sum is $\frac{1}{2}$ .

So we can find the probability of rolling $3$ or $11$ instead and just subtract that from $\frac{1}{2}$ , which seems easier. Without writing out a table, we can see that there are two ways to make $3$ , and two ways to make $11$ , for a probability of $\frac{4}{36}$ .

$\frac{1}{2}-\frac{4}{36}=\frac{14}{36}=\boxed{\frac{7}{18}}$ .

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2008 AMC 10B Problems/Problem 20

Contents

Problem

Solution 1

Solution 2

See also