AMC 8 · 2016 · #13

Grade 7 probabilitycounting

Archetype Overlap / Complement Counting └ Sub-archetype At-Least-One Complement: 1 − P(None)

probability-basiccombinations-basicsystematic-enumeration systematic-enumerationcomplementary-counting ↑ Prerequisites: combinations-basicprobability-basic

📏 Short solution 💡 2 insights

Problem

Pick two different numbers at random from {-2, -1, 0, 3, 4, 5} and multiply them. What is the probability that the product equals 0?

Pick an answer.

(A)

$dfrac{1}{6}$

(B)

$dfrac{1}{5}$

(C)

$dfrac{1}{4}$

(D)

$dfrac{1}{3}$

(E)

$dfrac{1}{2}$

AMC 8 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

The set is small (6 numbers), so Tool #2 (Systematic List) can enumerate every unordered pair without any formula — easy to count favorable pairs and total pairs by hand. The favorable event collapses to a clean condition: "the pair contains 0," because 0 times anything is 0 and none of the other numbers multiply to 0. Tool #16 (Complement) gives a one-line sanity check: P(product = 0) = 1 - P(0 is NOT chosen), so we can confirm the answer from the opposite direction.

1STEP 1

List every unordered pair of two different numbers, pairing each number only with those after it so none is missed or repeated.

{-2,-1}, {-2,0}, {-2,3}, {-2,4}, {-2,5}, {-1,0}, {-1,3}, {-1,4}, {-1,5}, {0,3}, {0,4}, {0,5}, {3,4}, {3,5}, {4,5}

2STEP 2

Count the pairs in the list — the sample space has 15 pairs in all.

5 + 4 + 3 + 2 + 1 = 15 pairs

3STEP 3

A product is 0 only when a factor is 0, and only 0 qualifies — so the favorable pairs are the 5 pairs containing 0.

{-2,0}, {-1,0}, {0,3}, {0,4}, {0,5} → 5 pairs

4STEP 4

Divide favorable by total and reduce: the probability is $\frac{1}{3}$ , choice (D).

P(product = 0) =

\frac{5}{15}

\frac{1}{3}

→ (D)

Answer

dfrac{1}{3}

There are 6 numbers and only 1 of them is 0. Picking 2 numbers out of 6, the chance that the special number 0 is in the pair is

\frac{2}{6}

\frac{1}{3}

— exactly choice (D). The size of the answer is sensible: 0 is one of six numbers, so it should show up about a third of the time when two are picked.

💡Key takeaway

List the pairs once, find the ones with 0, and divide — Grade 7 probability is enough to crack this AMC 8 problem.

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