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 "2002 AMC 8 Problems/Problem 21"
(→‎Solution 2 (fastest))
Line 16: Line 16:
 
We want the probability of at least two heads out of <math>4</math>. We can do this a faster way by noticing that the probabilities are symmetric around two heads.
 
We want the probability of at least two heads out of <math>4</math>. We can do this a faster way by noticing that the probabilities are symmetric around two heads.
 
Define <math>P(n)</math> as the probability of getting <math>n</math> heads on <math>4</math> rolls. Now our desired probability is <math>\frac{1-P(2)}{2} +P(2)</math>.
 
Define <math>P(n)</math> as the probability of getting <math>n</math> heads on <math>4</math> rolls. Now our desired probability is <math>\frac{1-P(2)}{2} +P(2)</math>.
−
We can easily calculate <math>P(2)</math> because there are <math>\binom{4}{2} = 6</math> ways to get <math>2</math> heads and <math>2</math> tails, and there are <math>2^4=16</math> total ways to flip these coins, giving <math>P(2)=\frac{6}{16}=\frac{3}{8}</math>, so plugging this in gives us <math>\boxed{\text{(E)}\ \frac{11}{16}}</math>.
+
We can easily calculate <math>P(2)</math> because there are <math>\binom{4}{2} = 6</math> ways to get <math>2</math> heads and <math>2</math> tails, and there are <math>2^4=16</math> total ways to flip these coins, giving <math>P(2)=\frac{6}{16}=\frac{3}{8}</math>, and plugging this in gives us <math>\boxed{\text{(E)}\ \frac{11}{16}}</math>.
 
~chrisdiamond10
 
~chrisdiamond10
  

Revision as of 00:08, 12 February 2021

Problem

Harold tosses a coin four times. The probability that he gets at least as many heads as tails is

$\text{(A)}\ \frac{5}{16}\qquad\text{(B)}\ \frac{3}{8}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{5}{8}\qquad\text{(E)}\ \frac{11}{16}$

Solution

Case 1: There are two heads, two tails. The number of ways to choose which two tosses are heads is $\binom{4}{2} = 6$, and the other two must be tails.

Case 2: There are three heads, one tail. There are $\binom{4}{1} = 4$ ways to choose which of the four tosses is a tail.

Case 3: There are four heads, no tails. This can only happen $1$ way.

There are a total of $2^4=16$ possible configurations, giving a probability of $\frac{6+4+1}{16} = \boxed{\text{(E)}\ \frac{11}{16}}$.

Solution 2 (fastest)

We want the probability of at least two heads out of $4$. We can do this a faster way by noticing that the probabilities are symmetric around two heads. Define $P(n)$ as the probability of getting $n$ heads on $4$ rolls. Now our desired probability is $\frac{1-P(2)}{2} +P(2)$. We can easily calculate $P(2)$ because there are $\binom{4}{2} = 6$ ways to get $2$ heads and $2$ tails, and there are $2^4=16$ total ways to flip these coins, giving $P(2)=\frac{6}{16}=\frac{3}{8}$, and plugging this in gives us $\boxed{\text{(E)}\ \frac{11}{16}}$. ~chrisdiamond10

See Also

2002 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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=2002_AMC_8_Problems/Problem_21&oldid=145815"