AMC 8 · 2014 · #18

Grade 7 probability

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

probability-basiccombinations-basicsystematic-enumeration caseworksystematic-enumeration ↑ Prerequisites: probability-basiccombinations-basic

📏 Medium solution 💡 3 insights

Problem

Four babies are born, and each is equally likely to be a boy or a girl. Among the listed gender splits — all boys; all girls; exactly 2 boys and 2 girls; 3 of one gender and 1 of the other — which split has the highest probability?

Pick an answer.

(A)

All are boys

(B)

All are girls

(C)

2 are boys and 2 are girls

(D)

3 are the same gender and 1 is not

(E)

They all have the same probability

AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

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.

1STEP 1

Each of the four births is boy-or-girl, so the sample space holds 16 equally likely ordered outcomes.

2 × 2 × 2 × 2 = 2⁴ = 16

2STEP 2

List all sixteen in a fixed order — sorted by number of boys — so nothing is missed and nothing is double-counted.

0 boys: GGGG ; 1 boy: BGGG, GBGG, GGBG, GGGB ; 2 boys: BBGG, BGBG, BGGB, GBBG, GBGB, GGBB ; 3 boys: BBBG, BBGB, BGBB, GBBB ; 4 boys: BBBB

3STEP 3

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.

(A)

\frac{1}{16}

, (B)

\frac{1}{16}

, (C)

\frac{6}{16}

, (D)

\frac{8}{16}

4STEP 4

Choice (E) dies once two counts differ, and $\frac{8}{16}$ beats every other count — so the 3-1 split is the most likely.

\frac{8}{16}

＞

\frac{6}{16}

＞

\frac{1}{16}

→ (D)

Answer

3 are the same gender and 1 is not

The four probabilities add to

\frac{1}{16}

\frac{1}{16}

\frac{6}{16}

\frac{8}{16}

\frac{16}{16}

= 1, which confirms the 16 ordered outcomes are partitioned correctly into the four categories. It also matches intuition: extreme splits like "all boys" or "all girls" each need a specific sequence and are rare, while a 3-1 split happens whenever any single child breaks from the majority — and there are 4 choices for which child that is, for each majority gender, giving 8 ways.

💡Key takeaway

With 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.

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