Difference between revisions of "2002 AMC 12A Problems/Problem 16"
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<math>\text{(A)}\ 2/5 \qquad \text{(B)}\ 9/20 \qquad \text{(C)}\ 1/2 \qquad \text{(D)}\ 11/20 \qquad \text{(E)}\ 24/25</math> | <math>\text{(A)}\ 2/5 \qquad \text{(B)}\ 9/20 \qquad \text{(C)}\ 1/2 \qquad \text{(D)}\ 11/20 \qquad \text{(E)}\ 24/25</math> | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/8WrdYLw9_ns?t=381 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==Solution== | ==Solution== | ||
Line 30: | Line 35: | ||
=== Solution 2 === | === Solution 2 === | ||
− | We | + | 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 | ||
+ | |||
+ | ==Solution 3== | ||
+ | We have 5 cases, if Tina choose <math>1, 2, 3, 4,</math> or <math>5.</math> | ||
+ | |||
+ | The number of ways of choosing 2 numbers from <math>5</math> are <math>\binom{5}{2}</math>. | ||
+ | -------------------------- | ||
+ | Case 1: Tina chooses <math>1</math>. | ||
+ | |||
+ | In this case, since the numbers are distinct, Tina can choose <math>(1, 2), (1, 3), (1, 4),</math> or <math>(1, 5).</math> | ||
+ | |||
+ | If Tina chooses <math>1</math> and <math>2</math> which some to <math>3</math>, Sergio only has <math>10-3=7</math> choices. | ||
+ | |||
+ | Since the sum of the combined numbers increases by <math>1</math> every time for this specific case, Sergio has <math>1</math> less choice every time. | ||
+ | |||
+ | Therefore, the probability of this is <math>\frac{7+6+5+4}{10 \cdot \binom{5}{2}}</math>. | ||
+ | -------------------------- | ||
+ | If you do this over and over again you will see that you have <math>\frac{(7+6+5+4)+(7+5+4+3)+(6+5+3+2)+(5+4+3+1)+(4+3+2+1)}{10 \cdot \binom{5}{2}} = \frac{80}{100} = \frac{4}{5}</math> probability. | ||
+ | |||
+ | But since we overcounted by 2 (e.g. <math>(1, 2)</math> and <math>(2, 1)</math>) we need to divide by <math>2.</math> | ||
+ | |||
+ | Thus our answer is <math>\frac{4}{5} \div 2 = \boxed{\textbf{(A)} \frac{2}{5}}.</math> | ||
+ | |||
+ | ~mathboy282 | ||
+ | |||
+ | Note: I will add in all the cases soon, kind of busy today so yea. | ||
+ | |||
+ | === Solution 4 === | ||
+ | |||
+ | Assume Sergio chooses from <math>{2,3,\ldots,10}</math>. The probably of Tina getting a sum of <math>6+x</math> and <math>6-x</math> (<math>x \leq 4</math>) are equal due to symmetry. The probability of Sergio choosing numbers higher/lower than <math>6+x</math> is equal to him choosing numbers lower/higher than <math>6-x</math>. Therefore over all of Tina's sums, the probability of Sergio choosing a number higher is equal to the probability of choosing a number lower. | ||
+ | |||
+ | The probability that they get the same value is <math>1/9</math>, so the probability of Sergio getting a higher number is <math>\frac{(9-1)/2}{9} = \frac49</math>. | ||
+ | |||
+ | Sergio never wins when choosing <math>1</math> so the probability is <math>\frac49 \frac{9}{10} + (0)\frac{1}{10} = \boxed{\textbf{(A)} \frac{2}{5}}.</math> | ||
− | + | ~zeric | |
− | |||
==See Also== | ==See Also== |
Latest revision as of 17:41, 24 April 2021
- The following problem is from both the 2002 AMC 12A #16 and 2002 AMC 10A #24, so both problems redirect to this page.
Contents
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?
Video Solution
https://youtu.be/8WrdYLw9_ns?t=381
~ pi_is_3.14
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 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
Solution 3
We have 5 cases, if Tina choose or
The number of ways of choosing 2 numbers from are .
Case 1: Tina chooses .
In this case, since the numbers are distinct, Tina can choose or
If Tina chooses and which some to , Sergio only has choices.
Since the sum of the combined numbers increases by every time for this specific case, Sergio has less choice every time.
Therefore, the probability of this is .
If you do this over and over again you will see that you have probability.
But since we overcounted by 2 (e.g. and ) we need to divide by
Thus our answer is
~mathboy282
Note: I will add in all the cases soon, kind of busy today so yea.
Solution 4
Assume Sergio chooses from . The probably of Tina getting a sum of and () are equal due to symmetry. The probability of Sergio choosing numbers higher/lower than is equal to him choosing numbers lower/higher than . Therefore over all of Tina's sums, the probability of Sergio choosing a number higher is equal to the probability of choosing a number lower.
The probability that they get the same value is , so the probability of Sergio getting a higher number is .
Sergio never wins when choosing so the probability is
~zeric
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.