AMC 10 · 2006 · #21

Grade 7 probability

Archetype Arithmetic Expression Decomposition └ Sub-archetype Multi-Segment Path or Count Sum

probability-basicratio-proportionsystematic-enumeration identify-subproblems ↑ Prerequisites: probability-basic

📏 Long solution 💡 3 insights

Problem

A single die is weighted so that the chances of rolling 1,2,3,4,5,6 are in the ratio 1:2:3:4:5:6 — bigger faces are more likely, in exact proportion to the number. Two such identical dice are rolled. Find the probability that the two faces add up to 7.

Pick an answer.

(A)

$\frac{4}{63}$

(B)

$\frac{1}{8}$

(C)

$\frac{8}{63}$

(D)

$\frac{1}{6}$

(E)

$\frac{2}{7}$

AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

The ratio 1:2:3:4:5:6 does not name the probabilities directly, so Tool #4 (Introduce a Variable) supplies the missing scale: call the smallest probability p, write the six probabilities as p,2p,…,6p, and use the fact that they add to 1 to pin down p. Once each face has a known probability, Tool #2 (Make a Systematic List) sweeps out every face pair that totals 7 so none is missed and none double-counted. Tool #7 (Identify Subproblems) splits the job cleanly: first find the single-die probabilities, then find each pair's probability by multiplying, then add the pieces — a sum-of-products that collapses to one fraction.

1STEP 1

Name the scale with a variable

The ratio says every face chance is a multiple of one common piece: let p=P(1), so P(k)=kp for all six faces.

P(k)=k p for k=1,2,3,4,5,6

2STEP 2

Use total probability to solve for p

One die's six chances must total 1, so 21p=1, giving p=1/21 and P(k)=k/21.

p+2p+3p+4p+5p+6p=21p=1 → p=1/21, P(k)=k/21

3STEP 3

List every face pair that totals 7

Ordered pairs totalling 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1). Independence gives P(a,b)=ab/441.

P(a,b)=a/21·b/21=a b/441

4STEP 4

Add the pair probabilities and simplify

The numerators 6,10,12,12,10,6 sum to 56, so P=56/441=8/63, which is choice (C).

P(sum=7)=(6+10+12+12+10+6)/441=56/441=8/63 → (C)

Answer

8/63

A quick sanity check on size: with a fair die, a total of 7 has probability 6/36=1/6≈0.167. Here 8/63≈0.127 is a bit smaller, which fits — the weighting pushes rolls toward high faces like 6, and a high roll on one die needs a low roll (probability only 1/21 or 2/21) on the other to still total 7, so sevens become slightly harder. The value also sits right beside the closely spaced trap answer 4/63, which is exactly half of ours — the number you would get by counting only three pairs instead of all six ordered pairs, a reminder to count both orders.

💡Key takeaway

Turn a ratio into probabilities by giving it a variable and making them add to 1; then a compound total is the sum of every ordered way to make it, each way being the two independent chances multiplied.

Name the scale with a variable
Use total probability to solve for p
List every face pair that totals 7
Add the pair probabilities and simplify

Name the scale with a variable

Use total probability to solve for p

List every face pair that totals 7

Add the pair probabilities and simplify

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