1999 AHSME Problems/Problem 24

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?

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

Solution

There are ${6 \choose 2} = 15$ chords, and therefore ${15 \choose 4}$ ways how to choose four chords.

Each set of four points corresponds to exactly one convex quadrilateral, therefore there are ${6\choose 4}$ cases in which the four chords form a convex quadrilateral.

The resulting probability is $\frac {{6\choose 4}}{{15 \choose 4}} = \frac{6\cdot 5\cdot 4\cdot 3}{15\cdot 14\cdot 13\cdot 12} = \frac{1}{7\cdot 13} = \boxed{\frac{1}{91}}$ .

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

Page

Toolbox

Search

1999 AHSME Problems/Problem 24

Problem

Solution

See also