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 "2004 AMC 12A Problems/Problem 20"

(→‎Solution 1)
m (minor edits)
Line 4: Line 4:
 
<math>\text {(A)}\ \frac14 \qquad \text {(B)}\ \frac13 \qquad \text {(C)}\ \frac12 \qquad \text {(D)}\ \frac23 \qquad \text {(E)}\ \frac34</math>
 
<math>\text {(A)}\ \frac14 \qquad \text {(B)}\ \frac13 \qquad \text {(C)}\ \frac12 \qquad \text {(D)}\ \frac23 \qquad \text {(E)}\ \frac34</math>
  
−
__TOC__
 
 
== Solution ==
 
== Solution ==
 
=== Solution 1 ===  
 
=== Solution 1 ===  
Line 25: Line 24:
 
Then the area producing the desired result is 3/4. Since the area of the unit square is 1, the probability is <math>\frac 34</math>.
 
Then the area producing the desired result is 3/4. Since the area of the unit square is 1, the probability is <math>\frac 34</math>.
  
−
== See also ==
+
== See Also ==
 
{{AMC12 box|year=2004|ab=A|num-b=19|num-a=21}}
 
{{AMC12 box|year=2004|ab=A|num-b=19|num-a=21}}
  
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:59, 18 October 2020

Problem

Select numbers $a$ and $b$ between $0$ and $1$ independently and at random, and let $c$ be their sum. Let $A, B$ and $C$ be the results when $a, b$ and $c$, respectively, are rounded to the nearest integer. What is the probability that $A + B = C$?

$\text {(A)}\ \frac14 \qquad \text {(B)}\ \frac13 \qquad \text {(C)}\ \frac12 \qquad \text {(D)}\ \frac23 \qquad \text {(E)}\ \frac34$

Solution

Solution 1

Casework:

  1. $0 + 0 = 0$. The probability that $a < \frac{1}{2}$ and $b < \frac{1}{2}$ is $\left(\frac 12\right)^2 = \frac{1}{4}$. Notice that the sum $a+b$ ranges from $0$ to $1$ with a symmetric distribution across $a+b=c=\frac 12$, and we want $c < \frac 12$. Thus the chance is $\frac{\frac{1}{4}}2 = \frac 18$.
  2. $0 + 1 = 1$. The probability that $a < \frac 12$ and $b > \frac 12$ is $\frac 14$, but now $\frac{1}{2} < a+b = c < \frac 32$, which makes $C = 1$ automatically. Hence the chance is $\frac 14$.
  3. $1 + 0 = 1$. This is the same as the previous case.
  4. $1 + 1 = 2$. We recognize that this is equivalent to the first case.

Our answer is $2\left(\frac 18 + \frac 14 \right) = \frac 34 \Rightarrow \mathrm{(E)}$.

Solution 2

Use areas to deal with this continuous probability problem. Set up a unit square with values of $a$ on x-axis and $b$ on y-axis.

If $a + b < 1/2$ then this will work because $A = B = C = 0$. Similarly if $a + b > 3/2$ then this will work because in order for this to happen, $a$ and $b$ are each greater than $1/2$ making $A = B = 1$, and $C = 2$. Each of these triangles in the unit square has area of 1/8.

The only case left is when $C = 1$. Then each of $A$ and $B$ must be 1 and 0, in any order. These cut off squares of area 1/2 from the upper left and lower right corners of the unit square.

Then the area producing the desired result is 3/4. Since the area of the unit square is 1, the probability is $\frac 34$.

See Also

2004 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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=2004_AMC_12A_Problems/Problem_20&oldid=135301"