Difference between revisions of "2002 AMC 12A Problems/Problem 16"
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Scrabbler94 (talk | contribs) (→Solution 3: existing solution 3 does not justify the symmetry, and we can't say in general that just because E(Tina's sum) = 6, we can just look for the probability that Sergio's # is >6.) |
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=== Solution 3 === | === Solution 3 === | ||
− | + | We invoke some symmetry. Let <math>T</math> denote Tina's sum, and let <math>S</math> denote Sergio's number. Observe that, for <math>i = 2, 3, \ldots, 10</math>, <math>\text{Pr}(T=i) = \text{Pr}(T=12-i)</math>. | |
+ | |||
+ | If Tina's sum is <math>i</math>, then the probability that Sergio's number is larger than Tina's sum is <math>\frac{10-i}{10}</math>. Thus, the probability <math>P</math> is | ||
+ | |||
+ | <cmath>P = \text{Pr}(S>T) = \sum_{i=2}^{10} \text{Pr}(T=i) \times \frac{10-i}{10}</cmath> | ||
+ | |||
+ | Using the symmetry observation, we can also write the above sum as | ||
+ | <cmath> P = \sum_{i=2}^{10} \text{Pr}(T=12-i) \times \frac{10-i}{10} = \sum_{i=2}^{10} \text{Pr}(T=i) \times \frac{i-2}{10}</cmath> | ||
+ | where the last equality follows as we reversed the indices of the sum (by replacing <math>12-i</math> with <math>i</math>). Thus, adding the two equivalent expressions for <math>P</math>, we have | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | 2P &= \sum_{i=2}^{10} \text{Pr}(T=i) \times \left(\frac{10-i}{10} + \frac{i-2}{10}\right) \\ | ||
+ | &= \sum_{i=2}^{10} \text{Pr}(T=i) \times \frac{4}{5} \\ | ||
+ | &= \frac{4}{5} \sum_{i=2}^{10} \text{Pr}(T=i) \\ | ||
+ | &= \frac{4}{5} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Since this represents twice the desired probability, the answer is <math>P = \boxed{\textbf{(A)} \frac{2}{5}}</math>. -scrabbler94 | ||
==See Also== | ==See Also== |
Revision as of 23:56, 23 November 2019
- The following problem is from both the 2002 AMC 12A #16 and 2002 AMC 10A #24, so both problems redirect to this page.
Problem
Tina randomly selects two distinct numbers from the set , and Sergio randomly selects a number from the set . What is the probability that Sergio's number is larger than the sum of the two numbers chosen by Tina?
Solution
Solution 1
This is not too bad using casework.
Tina gets a sum of 3: This happens in only one way and Sergio can choose a number from 4 to 10, inclusive. There are 7 ways that Sergio gets a desirable number here.
Tina gets a sum of 4: This once again happens in only one way . Sergio can choose a number from 5 to 10, so 6 ways here.
Tina gets a sum of 5: This can happen in two ways and . Sergio can choose a number from 6 to 10, so ways here.
Tina gets a sum of 6: Two ways here and . Sergio can choose a number from 7 to 10, so here.
Tina gets a sum of 7: Two ways here and . Sergio can choose from 8 to 10, so ways here.
Tina gets a sum of 8: Only one way possible ). Sergio chooses 9 or 10, so 2 ways here.
Tina gets a sum of 9: Only one way . Sergio must choose 10, so 1 way.
In all, there are ways. Tina chooses two distinct numbers in ways while Sergio chooses a number in ways, so there are ways in all. Since , our answer is .
Solution 2
We want to find the average of the smallest possible chance of Sergio winning and the largest possible chance of Sergio winning. This is because the probability decreases linearly. The largest possibility of Sergio winning if Tina chooses a 1 and a 2. The chances of Sergio winning is then . The smallest possibility of Sergio winning is if Tina chooses a 4 and a 5. The chances of Sergio winning then is . The average of and is .
Solution 3
We invoke some symmetry. Let denote Tina's sum, and let denote Sergio's number. Observe that, for , .
If Tina's sum is , then the probability that Sergio's number is larger than Tina's sum is . Thus, the probability is
Using the symmetry observation, we can also write the above sum as where the last equality follows as we reversed the indices of the sum (by replacing with ). Thus, adding the two equivalent expressions for , we have
Since this represents twice the desired probability, the answer is . -scrabbler94
See Also
2002 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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 |
2002 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.