AMC 8 · 2014 · #18
Grade 7 probabilityProblem
Four children were born at City Hospital yesterday. Assume each child is equally likely to be a boy or a girl. Which of the following outcomes is most likely?
Pick an answer.
AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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?
Givens: $4$ children, each independent; $P(\text{boy}) = P(\text{girl}) = \tfrac{1}{2}$; Answer choices: (A) all boys, (B) all girls, (C) $2$ boys and $2$ girls, (D) $3$ same gender and $1$ different, (E) all equally likely
Unknowns: Which outcome (A)-(E) has the highest probability
Understand
Restated: 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?
Givens: $4$ children, each independent; $P(\text{boy}) = P(\text{girl}) = \tfrac{1}{2}$; Answer choices: (A) all boys, (B) all girls, (C) $2$ boys and $2$ girls, (D) $3$ same gender and $1$ different, (E) all equally likely
Plan
Primary tool: #2 Make a Systematic List
Secondary: #3 Eliminate Possibilities
With only $4$ children there are $2^4 = 16$ 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.
Execute — Answer: D
3.OA.A.1 Step 1 - Count the total number of ordered outcomes.
- Each of the $4$ births has $2$ possibilities, so the sample space size is $2 \times 2 \times 2 \times 2$.
💡 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.
7.SP.C.8 Step 2 - List the $16$ outcomes in a fixed order — sort by the number of boys, then alphabetically within each group.
- This is the systematic-list move from Tool #2 and guarantees we miss nothing and double-count nothing.
💡 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.7 Step 3 - Read the count of each answer choice straight off the list.
- (A) all boys = just BBBB.
- (B) all girls = just GGGG.
- (C) $2$ boys and $2$ girls = the $6$ strings in the "$2$ boys" row.
- (D) $3$ of one gender and $1$ of the other = the "$1$ boy" row ($4$ strings) plus the "$3$ boys" row ($4$ strings), $4 + 4 = 8$ in all.
💡 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.
7.SP.C.8 Step 4 - Now eliminate.
- (E) ("all equally likely") is killed instantly by the fact that (A) and (C) already have different probabilities ($1/16$ vs $6/16$).
- Among (A)-(D), the biggest fraction is $8/16$, so (D) wins.
💡 Multiple-choice elimination (Tool #3) closes the problem the moment one option's count beats every other option's count.
3.OA.A.1 Count the total number of ordered outcomes. Each of the $4$ births has $2$ possi 7.SP.C.8 List the $16$ outcomes in a fixed order — sort by the number of boys, then alpha 7.SP.C.7 Read the count of each answer choice straight off the list. (A) all boys = just 7.SP.C.8 Now eliminate. (E) ("all equally likely") is killed instantly by the fact that ( Review
Reasonableness: The four probabilities add to $\tfrac{1}{16} + \tfrac{1}{16} + \tfrac{6}{16} + \tfrac{8}{16} = \tfrac{16}{16} = 1$, 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.
Alternative: Tool #9 (Easier Related Problem) with $2$ children: outcomes BB, BG, GB, GG show "$1$ of each" beats "both same" $2$ vs $2$. Extending to $4$ children, the lopsided $3$-$1$ split still has more arrangements than the all-same case but more than the perfectly even case too, because $\binom{4}{1} + \binom{4}{3} = 4 + 4 = 8$ exceeds $\binom{4}{2} = 6$. Same answer (D).
CCSS standards used (min grade 7)
3.OA.A.1Interpret products of whole numbers (Computing the size of the sample space as $2 \times 2 \times 2 \times 2 = 16$ by multiplying the number of options at each independent birth.)7.SP.C.7Develop a probability model and use it to find probabilities of events (Treating the $16$ ordered outcomes as equally likely so that the probability of each answer-choice category equals (favorable outcomes) $/$ $16$.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and tree diagrams (Listing all $16$ four-child gender strings in a systematic order and reading off how many strings belong to each answer-choice category.)
⭐ 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.
⭐ 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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