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Difference between revisions of "2004 AMC 12A Problems/Problem 20"

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<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>
  
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== Solution ==
 
== Solution ==
 
=== Solution 1 ===  
 
=== Solution 1 ===  
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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>.
  
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== See also ==
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== Solution 3 (Alcumus) ==
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The conditions under which <math>A+B=C</math> are as follows.
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(i) If <math>a+b< 1/2</math>, then <math>A=B=C=0</math>.
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(ii) If <math>a\geq 1/2</math> and <math>b<1/2</math>, then <math>B=0</math> and <math>A=C=1</math>.
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(iii) If <math>a<1/2</math> and <math>b\geq 1/2</math>, then <math>A=0</math> and <math>B=C=1</math>.
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(iv) If <math>a+b\geq 3/2</math>, then <math>A=B=1</math> and <math>C=2</math>.
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These conditions correspond to the shaded regions of the graph shown. The combined area of those regions is 3/4, and the area of the entire square is 1, so the requested probability is <math>\boxed{3/4}</math>.
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<asy>
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unitsize(2cm);
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draw((1.1,0)--(0,0)--(0,1.1),linewidth(1));
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fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,gray(0.7));
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fill((0.5,0)--(0.5,0.5)--(0,0.5)--cycle,white);
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fill((0.5,0.5)--(1,0.5)--(0.5,1)--cycle,white);
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label("$a$",(1.1,0),E);
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label("$b$",(0,1.1),N);
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label("1",(1,0),S);
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label("1",(0,1),W);
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label("0",(0,0),SW);
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</asy>
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== 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]]
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{{MAA Notice}}

Latest revision as of 22:26, 28 December 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 $\frac 34$.

Solution 3 (Alcumus)

The conditions under which $A+B=C$ are as follows.


(i) If $a+b< 1/2$, then $A=B=C=0$.


(ii) If $a\geq 1/2$ and $b<1/2$, then $B=0$ and $A=C=1$.


(iii) If $a<1/2$ and $b\geq 1/2$, then $A=0$ and $B=C=1$.


(iv) If $a+b\geq 3/2$, then $A=B=1$ and $C=2$.

These conditions correspond to the shaded regions of the graph shown. The combined area of those regions is 3/4, and the area of the entire square is 1, so the requested probability is $\boxed{3/4}$.

[asy] unitsize(2cm); draw((1.1,0)--(0,0)--(0,1.1),linewidth(1)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,gray(0.7)); fill((0.5,0)--(0.5,0.5)--(0,0.5)--cycle,white); fill((0.5,0.5)--(1,0.5)--(0.5,1)--cycle,white); label("$a$",(1.1,0),E); label("$b$",(0,1.1),N); label("1",(1,0),S); label("1",(0,1),W); label("0",(0,0),SW); [/asy]

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

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