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 "2010 AMC 12A Problems/Problem 15"

m (→‎Solution)
(→‎See also)
 
(4 intermediate revisions by 3 users not shown)
Line 1: Line 1:
−
== Problem 15 ==
+
== Problem ==
−
A coin is altered so that the probability that it lands on heads is less than <math>\frac{1}{2}</math> and when the coin is flipped four times, the probaiblity of an equal number of heads and tails is <math>\frac{1}{6}</math>. What is the probability that the coin lands on heads?
+
A coin is altered so that the probability that it lands on heads is less than <math>\frac{1}{2}</math> and when the coin is flipped four times, the probability of an equal number of heads and tails is <math>\frac{1}{6}</math>. What is the probability that the coin lands on heads?
  
 
<math>\textbf{(A)}\ \frac{\sqrt{15}-3}{6} \qquad \textbf{(B)}\ \frac{6-\sqrt{6\sqrt{6}+2}}{12} \qquad \textbf{(C)}\ \frac{\sqrt{2}-1}{2} \qquad \textbf{(D)}\ \frac{3-\sqrt{3}}{6} \qquad \textbf{(E)}\ \frac{\sqrt{3}-1}{2}</math>
 
<math>\textbf{(A)}\ \frac{\sqrt{15}-3}{6} \qquad \textbf{(B)}\ \frac{6-\sqrt{6\sqrt{6}+2}}{12} \qquad \textbf{(C)}\ \frac{\sqrt{2}-1}{2} \qquad \textbf{(D)}\ \frac{3-\sqrt{3}}{6} \qquad \textbf{(E)}\ \frac{\sqrt{3}-1}{2}</math>
Line 9: Line 9:
 
The probability of flipping <math>2</math> heads and <math>2</math> tails is equal to the number of ways to flip it times the product of the probability of flipping each coin.
 
The probability of flipping <math>2</math> heads and <math>2</math> tails is equal to the number of ways to flip it times the product of the probability of flipping each coin.
  
 +
<cmath>\begin{align*}{4 \choose 2}x^2(1-x)^2 &= \frac{1}{6}\\
 +
6x^2(1-x)^2 &= \frac{1}{6}\\
 +
x^2(1-x)^2 &= \frac{1}{36}\\
 +
x(1-x) &= \pm\frac{1}{6}\end{align*}</cmath>
  
−
<math>{4 \choose 2}x^2(1-x)^2 = \frac{1}{6}</math>
+
As for the desired probability <math>x</math> both <math>x</math> and <math>1-x</math> are nonnegative, we only need to consider the positive root, hence
  
−
<math>6x^2(1-x)^2 = \frac{1}{6}</math>
+
<cmath>\begin{align*}x(1-x) &= \frac{1}{6}\\
 +
6x^2-6x+1&=0\end{align*}</cmath>
  
−
<math>x^2(1-x)^2 = \frac{1}{36}</math>
+
Applying the quadratic formula we get that the roots of this equation are <math>\frac{3\pm\sqrt{3}}{6}</math>. As the probability of heads is less than <math>\frac{1}{2}</math>, we get that the answer is <math>\boxed{\textbf{(D)}\ \frac{3-\sqrt{3}}{6}}</math>.
  
−
<math>x(1-x) = \pm\frac{1}{6}</math>
+
== See also ==
 +
{{AMC12 box|year=2010|num-b=14|num-a=16|ab=A}}
  
−
<math>6x^2-6x\pm1=0</math>
+
[[Category:Introductory Combinatorics Problems]]
−
 
+
{{MAA Notice}}
−
 
 
−
Applying the quadratic formula we get <math>\frac{3\pm\sqrt{3}}{6}</math> or <math>\frac{3\pm\sqrt{15}}{6}</math>. The only answer that is less than <math>\frac{1}{2}</math> (since probability of heads is less than tails) and greater than <math>0</math> is <math>\boxed{\textbf{(D)}\ \frac{3-\sqrt{3}}{6}}</math>.
 

Latest revision as of 21:47, 3 July 2013

Problem

A coin is altered so that the probability that it lands on heads is less than $\frac{1}{2}$ and when the coin is flipped four times, the probability of an equal number of heads and tails is $\frac{1}{6}$. What is the probability that the coin lands on heads?

$\textbf{(A)}\ \frac{\sqrt{15}-3}{6} \qquad \textbf{(B)}\ \frac{6-\sqrt{6\sqrt{6}+2}}{12} \qquad \textbf{(C)}\ \frac{\sqrt{2}-1}{2} \qquad \textbf{(D)}\ \frac{3-\sqrt{3}}{6} \qquad \textbf{(E)}\ \frac{\sqrt{3}-1}{2}$

Solution

Let $x$ be the probability of flipping heads. It follows that the probability of flipping tails is $1-x$.

The probability of flipping $2$ heads and $2$ tails is equal to the number of ways to flip it times the product of the probability of flipping each coin.

\begin{align*}{4 \choose 2}x^2(1-x)^2 &= \frac{1}{6}\\ 6x^2(1-x)^2 &= \frac{1}{6}\\ x^2(1-x)^2 &= \frac{1}{36}\\ x(1-x) &= \pm\frac{1}{6}\end{align*}

As for the desired probability $x$ both $x$ and $1-x$ are nonnegative, we only need to consider the positive root, hence

\begin{align*}x(1-x) &= \frac{1}{6}\\ 6x^2-6x+1&=0\end{align*}

Applying the quadratic formula we get that the roots of this equation are $\frac{3\pm\sqrt{3}}{6}$. As the probability of heads is less than $\frac{1}{2}$, we get that the answer is $\boxed{\textbf{(D)}\ \frac{3-\sqrt{3}}{6}}$.

See also

2010 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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 AMC 12 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=2010_AMC_12A_Problems/Problem_15&oldid=53810"