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"

(→‎Problem)
(→‎Solution)
Line 7: Line 7:
 
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>.
 
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 \left(\frac{1}{2}\right)^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>.
+
The probability of C occurring is <math>\dbinom{4}{2}\cdot \left(\frac{1}{2}\right)^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>
 
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>

Revision as of 13:07, 28 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

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

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 $\dbinom{4}{2}\cdot \left(\frac{1}{2}\right)^4 = \frac{3}{8}$. 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=66364"