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| − | ==Problem==
| + | #redirect [[2002 AMC 12A Problems/Problem 16]] |
| − | Tina randomly selects two distinct numbers from the set {1, 2, 3, 4, 5}, and Sergio randomly selects a number from the set {1, 2, ..., 10}. What is the probability that Sergio's number is larger than the sum of the two numbers chosen by Tina?
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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>
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| − | ==Solution==
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| − | This is not too bad using casework.
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| − | Tina gets a sum of 3: This happens in only one way (1,2) and Sergio can choose a number from 4 to 10, inclusive. There are 7 ways that Sergio gets a desirable number here.
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| − | Tina gets a sum of 4: This once again happens in only one way (1,3). Sergio can choose a number from 5 to 10, so 6 ways here.
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| − | Tina gets a sum of 5: This can happen in two ways (1,4) and (2,3). Sergio can choose a number from 6 to 10, so 2*5=10 ways here.
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| − | Tina gets a sum of 6: Two ways here (1,5) and (2,4). Sergio can choose a number from 7 to 10, so 2*4=8 here.
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| − | Tina gets a sum of 7: Two ways here (2,5) and (3,4). Sergio can choose from 8 to 10, so 2*3=6 ways here.
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| − | Tina gets a sum of 8: Only one way possible (3,5). Sergio chooses 9 or 10, so 2 ways here.
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| − | Tina gets a sum of 9: Only one way (4,5). Sergio must choose 10, so 1 way.
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| − | In all, there are <math>7+6+10+8+6+2+1=40</math> ways. Tina chooses two distinct numbers in <math>\binom{5}{2}=10</math> ways while Sergio chooses a number in 10 ways, so there are 10*10=100 ways in all. Since <math>\frac{40}{100}=\frac{2}{5}</math>, our answer is <math>\boxed{\text{(A)}\ 2/5}</math>.
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| − | ==See Also==
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| − | {{AMC10 box|year=2002|ab=A|num-b=23|num-a=25}}
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| − | [[Category:Introductory Combinatorics Problems]] | |
