Competition · AMC preparation · step 4 of 4
AMC 8 · 2014 · #18
Grade 7 probabilityPick an answer.
AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
With only 4 children there are 2⁴ = 16 ordered outcomes, small enough to list every one (Tool #2). Once we have the full sample space, we group the 16 outcomes by category and the answer is just "which group is biggest?" Tool #3 (Eliminate Possibilities) then sweeps the multiple-choice list: (E) dies as soon as we see two categories with different counts, and (A), (B), (C) fall when (D) turns out to have more outcomes than any of them.
Count all the outcomes
Each of the four births is boy-or-girl, so the sample space holds 16 equally likely ordered outcomes.
Multiplying the number of choices at each step is the Grade 3 "product as repeated grouping" idea — 4 independent 2-way choices make 16 ordered strings.
3.OA.A.1Make A Systematic ListList the sixteen outcomes
List all sixteen in a fixed order — sorted by number of boys — so nothing is missed and nothing is double-counted.
Listing every outcome of a compound event in an organized way is exactly the Grade 7 "sample space by organized list" standard.
7.SP.C.8Make A Systematic ListCount each choice
Read each choice off the list: all-boys and all-girls each have 1, the 2-2 split has 6, and the 3-1 split has 8.
Because all 16 ordered outcomes are equally likely, the probability of a category is just (outcomes in category) / 16 — the Grade 7 equally-likely probability model.
Since all 16 four-child strings are equally likely, each listed outcome's chance is just its share of those 16 strings, and the "3 of one kind and 1 of the other" group fills 8 of them while every other listed group fills at most 6.
▸ Why?
The four births are four independent 2-way choices, and because independent choices multiply, the options at each birth combine to 2 × 2 × 2 × 2 = 16 strings in all.
▸ Why?
The "3 of one kind" group splits into two cases that never overlap — three boys with one girl, or three girls with one boy — so its total is the two case-counts added: 4 + 4 = 8.
▸ Why?
In each case the whole string is pinned down by the single child who differs, and that odd child can be any one of the 4 positions, so each case holds exactly 4 strings.
▸ Why?
A set of four cannot be "three boys and one girl" and "three girls and one boy" at the same time, so the two cases share no string and their counts simply add.
▸ Why?
The only other group with many strings, "2 and 2," is fixed by choosing which 2 of the 4 children are boys, and there are only 6 such choices — fewer than 8.
Compare and eliminate
Choice (E) dies once two counts differ, and beats every other count — so the 3-1 split is the most likely.
Multiple-choice elimination (Tool #3) closes the problem the moment one option's count beats every other option's count.
7.SP.C.8Eliminate PossibilitiesWith only 16 outcomes to list, the most likely group is the one with the most arrangements — and a 3-and-1 split has 8 of them, more than any other choice.
- Count all the outcomes
- List the sixteen outcomes
- Count each choice
- Compare and eliminate
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