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>. | ||
− | == See | + | == 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}} | {{MAA Notice}} |
Latest revision as of 22:26, 28 December 2020
Problem
Select numbers and between and independently and at random, and let be their sum. Let and be the results when and , respectively, are rounded to the nearest integer. What is the probability that ?
Solution
Solution 1
- . The probability that and is . Notice that the sum ranges from to with a symmetric distribution across , and we want . Thus the chance is .
- . The probability that and is , but now , which makes automatically. Hence the chance is .
- . This is the same as the previous case.
- . We recognize that this is equivalent to the first case.
Our answer is .
Solution 2
Use areas to deal with this continuous probability problem. Set up a unit square with values of on x-axis and on y-axis.
If then this will work because . Similarly if then this will work because in order for this to happen, and are each greater than making , and . Each of these triangles in the unit square has area of 1/8.
The only case left is when . Then each of and 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 .
Solution 3 (Alcumus)
The conditions under which are as follows.
(i) If , then .
(ii) If and , then and .
(iii) If and , then and .
(iv) If , then and .
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 .
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.