Difference between revisions of "2012 AMC 10A Problems/Problem 25"
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Without loss of generality, assume that <math>0 \le x \le y \le z \le n</math>. | Without loss of generality, assume that <math>0 \le x \le y \le z \le n</math>. | ||
− | Then, the possible choices for <math>x</math>, <math>y</math>, and <math>z</math> | + | Then, the possible choices for <math>x</math>, <math>y</math>, and <math>z</math> are represented by the expression <math>\frac{n^3}{3!}=\frac{n^3}{6}</math>. |
There are two restrictions: <math>x+1 \le y</math> and <math>y+1 \le z</math>. | There are two restrictions: <math>x+1 \le y</math> and <math>y+1 \le z</math>. | ||
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Since <math>0 \le x</math>, then <math>2 \le x'</math>. Thus, <math>2 \le x' \le y' \le z \le n</math>. | Since <math>0 \le x</math>, then <math>2 \le x'</math>. Thus, <math>2 \le x' \le y' \le z \le n</math>. | ||
− | There are <math>n-2</math> ways to choose each number; the successful choices | + | There are <math>n-2</math> ways to choose each number; the successful choices are represented by <math>\frac{(n-2)^3}{6}</math>. |
− | The probability then, is <math>\text{P}=\frac{\text{successful}}{\text{possible}}=\frac{\frac{n^3}{6}}{\frac{ | + | The probability then, is <math>\text{P}=\frac{\text{successful}}{\text{possible}}=\frac{\frac{(n-2)^3}{6}}{\frac{n^3}{6}}</math> which must be greater than <math>\frac{1}{2}</math>. |
Plug in values. Trying <math>(\text{C})</math> gives us <math>\frac{343}{729}</math>, which is less than <math>\frac{1}{2}</math>. Try the next integer, <math>10</math>, which gives us <math>\frac{512}{1000}</math> which is greater than <math>\frac{1}{2}</math>. | Plug in values. Trying <math>(\text{C})</math> gives us <math>\frac{343}{729}</math>, which is less than <math>\frac{1}{2}</math>. Try the next integer, <math>10</math>, which gives us <math>\frac{512}{1000}</math> which is greater than <math>\frac{1}{2}</math>. |
Revision as of 01:00, 19 February 2012
Problem
Real numbers ,
, and
are chosen independently and at random from the interval
for some positive integer
. The probability that no two of
,
, and
are within 1 unit of each other is greater than
. What is the smallest possible value of
?
Solution
Without loss of generality, assume that .
Then, the possible choices for ,
, and
are represented by the expression
.
There are two restrictions: and
.
Now, let . Then,
and
.
Then, let . Combining the two inequalities gives us
.
Since , then
. Thus,
.
There are ways to choose each number; the successful choices are represented by
.
The probability then, is which must be greater than
.
Plug in values. Trying gives us
, which is less than
. Try the next integer,
, which gives us
which is greater than
.
Thus, our answer is .
See Also
2012 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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 10 Problems and Solutions |