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 "2015 AMC 12B Problems/Problem 17"
(→‎Solution)
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
−
When tossed <math>n</math> times, the probability of getting exactly 2 heads and the rest tails could be written as <math>{\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}</math>. However, we must account for the different orders that this could occur in (for example like HTT or THT or TTH). We can account for this by multiplying by <math>\dbinom{n}{2}</math>.
+
When tossed <math>n</math> times, the probability of getting exactly 2 heads and the rest tails is
  
−
Similarly, the probability of getting exactly 3 heads is <math>\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}</math>.
+
<cmath>\dbinom{n}{2} {\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}.</cmath>
 +
 
 +
Similarly, the probability of getting exactly 3 heads is
 +
 
 +
<cmath>\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}.</cmath>
  
 
Now set the two probabilities equal to each other and solve for <math>n</math>:
 
Now set the two probabilities equal to each other and solve for <math>n</math>:

Revision as of 00:27, 12 March 2015

Problem

An unfair coin lands on heads with a probability of $\tfrac{1}{4}$. When tossed $n$ times, the probability of exactly two heads is the same as the probability of exactly three heads. What is the value of $n$ ?

$\textbf{(A)}\; 5 \qquad\textbf{(B)}\; 8 \qquad\textbf{(C)}\; 10 \qquad\textbf{(D)}\; 11 \qquad\textbf{(E)}\; 13$

Solution

When tossed $n$ times, the probability of getting exactly 2 heads and the rest tails is

\[\dbinom{n}{2} {\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}.\]

Similarly, the probability of getting exactly 3 heads is

\[\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}.\]

Now set the two probabilities equal to each other and solve for $n$:

\[\dbinom{n}{2}{\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}=\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}\]

\[\frac{n(n-1)}{2!} \cdot \frac{3}{4} = \frac{n(n-1)(n-2)}{3!} \cdot \frac{1}{4}\]

\[3 = \frac{n-2}{3}\]

\[n-2 = 9\]

\[n = \fbox{\textbf{(D)}\; 11}\]

See Also

2015 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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=2015_AMC_12B_Problems/Problem_17&oldid=68763"