Difference between revisions of "1999 AHSME Problems/Problem 24"
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<math> \mathrm{(A) \ } \frac 1{15} \qquad \mathrm{(B) \ } \frac 1{91} \qquad \mathrm{(C) \ } \frac 1{273} \qquad \mathrm{(D) \ } \frac 1{455} \qquad \mathrm{(E) \ } \frac 1{1365}</math> | <math> \mathrm{(A) \ } \frac 1{15} \qquad \mathrm{(B) \ } \frac 1{91} \qquad \mathrm{(C) \ } \frac 1{273} \qquad \mathrm{(D) \ } \frac 1{455} \qquad \mathrm{(E) \ } \frac 1{1365}</math> | ||
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| + | == Solution == | ||
| + | There are 15 chords on the circle, which is found by using the diagonal formula n(n-3)/2 where n is 6, and adding 6 adjacent pairs of points. There are 15 choose 4 ways to choose chords, which is 15*14*13*12 divided by 4*3*2*1 = 91 * 15. Choosing four chords that form a convex quadrilateral is the same as choosing four points, since there is only one convex quadrilateral that can be formed with those points. So, the favorable outcomes is 6 Choose 4 which equals 6 Choose 2 which equals 6*5 divided by 2 which equals 15. Therefore, the answer is 15/91*15 = 1/91 = Option B | ||
== See also == | == See also == | ||
Latest revision as of 16:35, 2 February 2025
Problem
Six points on a circle are given. Four of the chords joining pairs of the six points are selected at random. What is the probability that the four chords form a convex quadrilateral?
Solution
There are 15 chords on the circle, which is found by using the diagonal formula n(n-3)/2 where n is 6, and adding 6 adjacent pairs of points. There are 15 choose 4 ways to choose chords, which is 15*14*13*12 divided by 4*3*2*1 = 91 * 15. Choosing four chords that form a convex quadrilateral is the same as choosing four points, since there is only one convex quadrilateral that can be formed with those points. So, the favorable outcomes is 6 Choose 4 which equals 6 Choose 2 which equals 6*5 divided by 2 which equals 15. Therefore, the answer is 15/91*15 = 1/91 = Option B
See also
| 1999 AHSME (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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
