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

m (See also: grrr...)
(Solution 3 (Alcumus))
 
(5 intermediate revisions by 4 users not shown)
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 ==
+
== Solution 3 (Alcumus) ==
 +
The conditions under which <math>A+B=C</math> are as follows.
 +
 
 +
 
 +
(i) If <math>a+b< 1/2</math>, then <math>A=B=C=0</math>.
 +
 
 +
 
 +
(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>.
 +
 
 +
 
 +
(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>.
 +
 
 +
 
 +
(iv) If <math>a+b\geq 3/2</math>, then <math>A=B=1</math> and <math>C=2</math>.
 +
 
 +
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>.
 +
 
 +
<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 ==
 
{{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}}

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 (ProblemsAnswer KeyResources)
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

Invalid username
Login to AoPS