Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2014 AMC 8 Problems/Problem 18"

(→‎Solution)
(→‎Solution)
Line 4: Line 4:
 
<math> \textbf{(A) }</math> all <math>4</math> are boys <math>\qquad\textbf{(B) }</math> all <math>4</math> are girls <math>\qquad\textbf{(C) }</math> <math>2</math> are girls and <math>2</math> are boys <math>\qquad\textbf{(D) }</math> <math>3</math> are of one gender and <math>1</math> is of the other gender <math>\qquad\textbf{(E) }</math> all of these outcomes are equally likely  
 
<math> \textbf{(A) }</math> all <math>4</math> are boys <math>\qquad\textbf{(B) }</math> all <math>4</math> are girls <math>\qquad\textbf{(C) }</math> <math>2</math> are girls and <math>2</math> are boys <math>\qquad\textbf{(D) }</math> <math>3</math> are of one gender and <math>1</math> is of the other gender <math>\qquad\textbf{(E) }</math> all of these outcomes are equally likely  
 
==Solution==
 
==Solution==
−
We'll just start by breaking cases down. The probability of A occurring is <math>(\frac{1}{2})^4 = \frac{1}{16}</math>. The probability of B occurring is <math>(\frac{1}{2})^4 = \frac{1}{16}</math>. The probability of C occurring is <math>\frac{4!}{2!2!}*(\frac{1}{2})^4 = \frac{3}{8}</math>. Lastly, the probability of D occurring is <math>2*\frac{4!}{3!}*(\frac{1}{2})^4 = \frac{1}{2}</math>. So out of the four fractions, D is the largest. So our answer is D.
+
We'll just start by breaking cases down. The probability of A occurring is <math>\left(\frac{1}{2}\right)^4 = \frac{1}{16}</math>. The probability of B occurring is <math>\left(\frac{1}{2}\right)^4 = \frac{1}{16}</math>. The probability of C occurring is <math>\frac{4!}{2!2!}\cdot (\frac{1}{2})^4 = \frac{3}{8}</math>. Lastly, the probability of D occurring is <math>2\cdot \frac{4!}{3!}\cdot\left(\frac{1}{2}\right)^4 = \frac{1}{2}</math>. So out of the four fractions, D is the largest. So our answer is <math>\boxed{\text{(D) 3 of one gender and 1 of the other}}.</math>
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2014|num-b=17|num-a=19}}
 
{{AMC8 box|year=2014|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:04, 27 November 2014

Problem

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

$\textbf{(A) }$ all $4$ are boys $\qquad\textbf{(B) }$ all $4$ are girls $\qquad\textbf{(C) }$ $2$ are girls and $2$ are boys $\qquad\textbf{(D) }$ $3$ are of one gender and $1$ is of the other gender $\qquad\textbf{(E) }$ all of these outcomes are equally likely

Solution

We'll just start by breaking cases down. The probability of A occurring is $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$. The probability of B occurring is $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$. The probability of C occurring is $\frac{4!}{2!2!}\cdot (\frac{1}{2})^4 = \frac{3}{8}$. Lastly, the probability of D occurring is $2\cdot \frac{4!}{3!}\cdot\left(\frac{1}{2}\right)^4 = \frac{1}{2}$. So out of the four fractions, D is the largest. So our answer is $\boxed{\text{(D) 3 of one gender and 1 of the other}}.$

See Also

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_8_Problems/Problem_18&oldid=66343"