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

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:

$0 + 0 = 0$ . 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 $\frac{\frac{1}{4}}2 = \frac 18$ .
$0 + 1 = 1$ . 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 $\frac 14$ .
$1 + 0 = 1$ . This is the same as the previous case.
$1 + 1 = 2$ . 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$",(0,1.1),N); label("1",(1,0),S); label("1",(0,1),W); label("0",(0,0),SW); [/asy]$

2004 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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