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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 ==
 
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There are <math>{6 \choose 2} = 15</math> chords, and therefore <math>{15 \choose 4}</math> ways how to choose four chords.
 
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Each set of four points corresponds to exactly one convex quadrilateral, therefore there are <math>{6\choose 4}</math> cases in which the four chords form a convex quadrilateral.
 
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The resulting probability is <math>\frac {{6\choose 4}}{{15 \choose 4}} = \frac{6\cdot 5\cdot 4\cdot 3}{15\cdot 14\cdot 13\cdot 12} = \frac{1}{7\cdot 13} = \boxed{\frac{1}{91}}</math>.
 
  
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1999|num-b=23|num-a=25}}
 
{{AHSME box|year=1999|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:57, 3 March 2024

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?

$\mathrm{(A) \ } \frac 1{15} \qquad \mathrm{(B) \ } \frac 1{91} \qquad \mathrm{(C) \ } \frac 1{273} \qquad \mathrm{(D) \ } \frac 1{455} \qquad \mathrm{(E) \ } \frac 1{1365}$

See also

1999 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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