AMC 10 · 2003 · #22
Grade 7 probabilityPick an answer.
The event 'they meet' looks like it stretches over an unbounded number of turns, so the first job is to shrink it. Tool #4 (Introduce a Variable) supplies the quantity s=x+y, which rises by exactly 1 for A and falls by exactly 1 for B every turn; that single variable proves the meeting can only happen on turn 6 and nowhere else. Tool #7 (Identify Subproblems) then splits that one turn into a handful of disjoint cases, one per possible meeting point, so the probability becomes a sum instead of a tangle. Tool #1 (Draw a Diagram) locates those points on the line x+y=6 and shows which of them B can actually reach. Tool #2 (Make a Systematic List) counts the six-step routes to each point using the addition rule, with the row total 64=2⁶ as a built-in audit. Tool #15 (Organize Information in More Ways) reappears in the review, where re-reading a pair of meeting routes as one long route recovers the same count in a single stroke. Tool #3 (Eliminate Possibilities) does the last, cheap job of matching the exact fraction against five decimals.
Turn x+y into a clock
The coordinate sum acts as a clock: it rises for one object and falls for the other, forcing a meeting at turn 6.
Every turn nudges A one rung up a diagonal ladder and B one rung down it, so there is exactly one rung where they can be level.
6.EE.A.2Introduce A VariableList the reachable rung-6 points
Matching coordinates at that turn leaves exactly six possible meeting points.
Only a point both objects can reach in six moves is a candidate, and matching the x-coordinates names each candidate exactly once.
6.NS.C.8Draw A DiagramCount six-step routes
Six-step routes are counted by the binomial row 1,6,15,20,15,6,1, which sums to 64.
A route is just a six-letter word, so counting routes to a point is counting words that use one letter a fixed number of times.
4.OA.C.5Make A Systematic ListConvert counts into equally likely outcomes
Independence makes every route pair equally likely, giving 4096 outcomes.
Twelve independent fair choices produce 4096 equally likely stories, so the whole task reduces to counting the good ones.
Twelve independent fair choices produce 4096 equally likely stories, so the whole task is counting good ones.
▸ Why?
One move tells you nothing about another, so the chance of a whole route is the individual chances multiplied.
▸ Why?
Because every story carries the same weight, the probability is just the good count over the total count.
Add the six disjoint cases
Adding the six disjoint cases gives 792 favourable pairs, or 99/512.
Each meeting point contributes A's ways times B's ways, and because the points never overlap the contributions simply add.
4.OA.A.3Make A Systematic ListRound and choose
That is about 0.1934, closest to 0.20, choice (C).
The exact fraction sits just under one fifth, and only one choice comes within a hundredth of it.
6.NS.B.3Eliminate PossibilitiesAdd the two coordinates to make a clock: it tells you the one turn on which the walkers can possibly be level, and after that the problem is just counting six-letter words.
- Turn x+y into a clock
- List the reachable rung-6 points
- Count six-step routes
- Convert counts into equally likely outcomes
- Add the six disjoint cases
- Round and choose