Competition · AMC preparation · step 4 of 4
AMC 8 · 2013 · #14
Grade 7 probabilitycountingPick an answer.
AMC 8 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
A match can happen in two completely separate ways — both pick green, or both pick red — so Tool #7 (Identify Subproblems) lets us solve each case on its own and add the results, since the two cases cannot happen at the same time. Tool #4 (Make a Systematic List) gives a useful cross-check: there are 2 × 4 = 8 equally likely (Abe-pick, Bob-pick) pairs, and we can just count how many of those 8 pairs have matching colors.
List the matching colors
A match needs the color in both hands. Abe has {green, red}, Bob {green, yellow, red}, so only green and red can match — two cases.
Listing the sample space and ruling out impossible outcomes is exactly the Grade 7 "find probabilities of compound events" move.
7.SP.C.8Identify SubproblemsFind the chance of two greens
Abe picks green with chance , Bob with . Picks are independent, so multiply: P(both green) = .
For independent events, the chance of both happening is the product of the two chances — the Grade 7 multiplication rule.
The chance that Abe and Bob both pick a green jelly bean equals Abe's chance of picking green times Bob's chance of picking green.
▸ Why?
Abe's draw and Bob's draw do not change each other, so each combined pick is one entry in a fixed list of equally likely outcomes; the chance of 'both green' is just how many entries are both-green out of all the entries, and multiplying the two separate chances is a way to count exactly that fraction.
▸ Why?
Lay the outcomes in a grid — Abe's 2 beans across the top, Bob's 4 beans down the side — which is 2 equal rows of 4, giving 8 pairs in all, and the single pair where Abe's green meets Bob's green is the only 'both green' outcome.
▸ Why?
The grid holds 8 pairs because 2 rows that each split into 4 columns is 2 equal groups of 4.
▸ Why?
Because no bean is favored over any other, all 8 pairs are equally likely, so the chance of 'both green' is its 1 favorable pair divided by the 8 total — exactly the 1/8 that 1/2 × 1/4 produces.
Find the chance of two reds
Abe picks red with chance , Bob with = (two red beans). Multiply: P(both red) = .
Same multiplication rule, applied to the second subproblem.
7.SP.C.8Identify SubproblemsAdd the two chances
The two cases can't both happen, so add them: + = + = , choice (C).
Adding fractions with unlike denominators by rewriting them with a common denominator is the Grade 5 fraction-addition standard.
5.NF.A.1Identify SubproblemsThis AMC 8 problem only needs the Grade 7 idea that you multiply chances for independent picks and add chances for cases that can't both happen!
- List the matching colors
- Find the chance of two greens
- Find the chance of two reds
- Add the two chances
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