Problem 18 · 2014 AMC 8
Medium
Counting & Probability
binomial-counting
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?
Show answer
Answer: D — 3 of one gender and 1 of the other is most likely.
Show hints
Hint 1 of 2
Outcomes feel unequal because they're not single sequences — the fair coins are the 16 ordered strings BBBB…GGGG. Count how many strings each description sweeps up.
Still stuck? Show hint 2 →
Hint 2 of 2
Watch the hidden lump: "3 of one gender and 1 of the other" secretly covers BOTH 3 boys-1 girl AND 3 girls-1 boy, so it gathers twice as many strings as you'd guess.
Show solution
Approach: count favorable sequences out of 16
- All 16 birth orders are equally likely; count how many each option covers. All 4 boys: 1 string. All 4 girls: 1 string.
- 2-and-2: choose which 2 of the 4 are girls, C(4, 2) = 6 strings.
- 3-and-1: this lumps BOTH directions — C(4, 1) (one girl) + C(4, 3) (three girls) = 4 + 4 = 8 strings.
- Largest count is 3 of one gender, 1 of the other with 8 strings (and 1 + 1 + 6 + 8 = 16 — check, all cases accounted for).
- Why this transfers: "most likely" means "covers the most equally-likely outcomes," and an option phrased as "k of one, the rest of the other" quietly bundles two binomial counts. The counts 1, 4, 6, 4, 1 are row 4 of Pascal's triangle.
Mark:
· log in to save