AMC 10 · 2008 · #20

Grade 7 probability
probability-basicparitycomplementary-counting complementary-counting ↑ Prerequisites: probability-basic
📏 Medium solution 💡 3 insights
Problem
One cube has faces 1, 2, 2, 3, 3, 4; the other has faces 1, 3, 4, 5, 6, 8. Roll both and add the two top numbers. Find the probability that this sum is 5, 7, or 9.

Pick an answer.

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

AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

The clean way through a two-dice problem is a full addition table (tool #2): rows are Die A's six faces, columns are Die B's six faces, and each cell holds the sum. That lays out all 36 equally likely outcomes at once so nothing is missed or double-counted. Tool #7 (Identify Subproblems) then splits the count into three easy tallies — how many cells equal 5, how many equal 7, how many equal 9 — and adds them. Tool #3 (Eliminate Possibilities) matches the reduced fraction to a choice and rules out the decoys. Tool #16 (Count the Complement) gives a fast independent sanity check via the complementary sums.

1STEP 1

Set up the equally likely outcomes

Both dice are fair and independent, so all 6 × 6 = 36 face pairs are equally likely — keep the repeated faces as separate outcomes.

6 × 6 = 36 equally likely face pairs
2STEP 2

Build the addition table

Put Die A's faces down the side and Die B's across the top, then fill each cell with its row-plus-column sum — all 36 sums at once.

+ & 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
3STEP 3

Count the favorable cells

Sweep the table once per target: sum 5 fills 4 cells, sum 7 fills 6, sum 9 fills 4, so the favorable count is 14.

4 (sum 5) + 6 (sum 7) + 4 (sum 9) = 14
4STEP 4

Form and reduce the probability

The probability is favorable over total, 14/36; dividing top and bottom by the common factor 2 gives 7/18.

14/36 = 7/18
5STEP 5

Match the answer choice

7/18 is choice (B): 11/18 is the complement, 5/18 drops the sum-7 cells, and 3/4, 8/9 are far too large.

7/18 → (B)
Answer
7/18
The three tallies 4, 6, 4 sum to 14 favorable out of 36, giving 14/36 = 7/18. This is between 0 and 1 and slightly below 1/2, which is reasonable for three chosen sums. As a check, notice both dice can only produce sums from 2 to 12, and the odd targets 5, 7, 9 sit right in the busy middle of the range, so a count near a third of the outcomes is believable.
💡Key takeaway

Make an addition table of all 36 face pairs, count the 14 cells that add to 5, 7, or 9, and the probability is 14/36 = 7/18.

  • Set up the equally likely outcomes
  • Build the addition table
  • Count the favorable cells
  • Form and reduce the probability
  • Match the answer choice