Difference between revisions of "2004 AMC 12A Problems/Problem 20"
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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>. | 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); | unitsize(2cm); | ||
draw((1.1,0)--(0,0)--(0,1.1),linewidth(1)); | draw((1.1,0)--(0,0)--(0,1.1),linewidth(1)); | ||
Line 47: | Line 47: | ||
fill((0.5,0)--(0.5,0.5)--(0,0.5)--cycle,white); | 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); | fill((0.5,0.5)--(1,0.5)--(0.5,1)--cycle,white); | ||
− | label(" | + | label("$a$",(1.1,0),E); |
− | label(" | + | label("$b$",(0,1.1),N); |
label("1",(1,0),S); | label("1",(1,0),S); | ||
label("1",(0,1),W); | label("1",(0,1),W); | ||
label("0",(0,0),SW); | label("0",(0,0),SW); | ||
− | + | </asy> | |
== See Also == | == See Also == |
Latest revision as of 23:26, 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.