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
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 · 1992 AMC 8 · 1993 AMC 8 · 1994 AMC 8 · 1995 AMC 8 · 1996 AMC 8 · 1997 AMC 8 · 1998 AMC 8 · 1999 AMC 8 · 2000 AMC 8 · 2001 AMC 8 · 2002 AMC 8 · 2003 AMC 8 · 2004 AMC 8 · 2005 AMC 8 · 2006 AMC 8 · 2007 AMC 8 · 2008 AMC 8 · 2009 AMC 8 · 2010 AMC 8 · 2011 AMC 8 · 2012 AMC 8 · 2013 AMC 8 · 2015 AMC 8 · 2016 AMC 8 · 2017 AMC 8 · 2018 AMC 8 · 2019 AMC 8 · 2020 AMC 8 · 2022 AMC 8 · 2023 AMC 8 · 2024 AMC 8 · 2025 AMC 8 · 2026 AMC 8