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?
$\mathrm{(A) \ } \frac 1{15} \qquad \mathrm{(B) \ } \frac 1{91} \qquad \mathrm{(C) \ } \frac 1{273} \qquad \mathrm{(D) \ } \frac 1{455} \qquad \mathrm{(E) \ } \frac 1{1365}$
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 $\frac {{6\choose 4}}{{15 \choose 4}} = \frac{6\cdot 5\cdot 4\cdot 3}{15\cdot 14\cdot 13\cdot 12} = \frac{1}{7\cdot 13} = \boxed{\frac{1}{91}}$.