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2004 AMC 12A Problems/Problem 20

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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 ?

\text {(A)}\ \frac14 \qquad \text {(B)}\ \frac13 \qquad \text {(C)}\ \frac12 \qquad \text {(D)}\ \frac23 \qquad \text {(E)}\ \frac34

Contents

Solution

Solution 1

Casework:

  1. . The probability that and is \left(\frac 12\right)^2 = \frac{1}{4}. Notice that the sum ranges from to with a symmetric distribution across , and we want . Thus the chance is .
  2. . The probability that and is , but now \frac{1}{2} < a+b = c < \frac 32, which makes automatically. Hence the chance is .
  3. . This is the same as the previous case.
  4. . 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 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 .

See also

2004 AMC 12A (Problems)
Preceded by
Problem 19
Followed by
Problem 21
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