2014 AMC 8 Problem 18
Attempt Problem 18 of the 2014 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2014 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
18.
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?
all are boys
all are girls
are girls and are boys
are of one gender and is of the
other gender
all of these outcomes are equally likely
Answer: D
Small hint:
List outcomes by the number of boys and girls.
Big hint:
The most likely category is the one with the most arrangements.
Solution:
There are equally likely birth-order outcomes. The counts by category are: all boys, ; all girls, ; two boys and two girls, ; and three of one gender and one of the other, .
The largest count is , so the most likely outcome is three children of one gender and one of the other.
Thus, D is the correct answer.
Problem 18 in other years
- 1985 AMC 8
- 1986 AMC 8
- 1987 AMC 8
- 1988 AMC 8
- 1989 AMC 8
- 1990 AMC 8
- 1991 AMC 8
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- 1993 AMC 8
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