Difference between revisions of "2024 AMC 12B Problems/Problem 17"
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(6) <math> (x_1,x_2,x_3) </math> = (-1,-2,-3) , b = 11 invalid | (6) <math> (x_1,x_2,x_3) </math> = (-1,-2,-3) , b = 11 invalid | ||
| − | probability = <math>\frac{4}{21 | + | the total event space is 21 (choice of select a) <math>\cdot</math> (21- 1) (choice of selecting b given no-replacement) |
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| + | hence, probability = <math>\frac{4}{21 \cdot 20}</math> = <math>\boxed{\textbf{(C) }\frac{1}{105}}</math> | ||
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
Revision as of 22:29, 14 November 2024
Problem 17
Integers 
and 
are randomly chosen without replacement from the set of integers with absolute value not exceeding 
. What is the probability that the polynomial 
has 
distinct integer roots?
.
Solution 1
-10 
a, b 10 , each of a,b has 21 choices
Applying Vieta,
Case:
(1) 
= (-1,-1,-6) , b = 13 invalid
(2) = (-1,1,6) , b = -1, a=-6 valid
(3) = ( 1,2,-3) , b = -7, a=0 valid
(4) = (1,-2,3) , b = -7, a=2 valid
(5) = (-1,2,3) , b = 1, a=4 valid
(6) = (-1,-2,-3) , b = 11 invalid
the total event space is 21 (choice of select a) 
(21- 1) (choice of selecting b given no-replacement)
hence, probability = 
= 
See also
| 2024 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 16 |
Followed by Problem 18 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
