Four points $A_1,$ $A_2,$ $A_3,$ and $A_4$ are chosen at random on a circle. Find the probability that chords $\overline{A_1 A_2}$ and $\overline{A_3 A_4}$ intersect.
Out of 4 ways, there is only one way to set up the chords so that A_1 A_2 and A_3 A_4 intersect, so the probability is 1/4.
+26387 Four points \(A_1,\ A_2,\ A_3,\ \)and \(A_4\) are chosen at random on a circle.
Find the probability that chords \(\overline{A_1 A_2}\) and \(\overline{A_3 A_4}\) intersect.
My attempt:
\(\begin{array}{|r|cl|} \hline & & \text{intersect} \\ \hline & 2~3 \\ 1 & 1~~~4 \\ \hline & 2~4 \\ 2 & 1~~~3 \\ \hline & 3~2 \\ 3 & 1~~~4& \checkmark\\ \hline & 3~4 \\ 4 & 1~~~2 \\ \hline & 4~2 \\ 5 & 1~~~3 & \checkmark\\ \hline & 4~3 \\ 6 & 1~~~2 \\ \hline & 1~3 \\ 7 & 2~~~4 \\ \hline & 1~4 \\ 8 & 2~~~3 \\ \hline & 3~1 \\ 9 & 2~~~4 & \checkmark\\ \hline & 3~4 \\ 10 & 2~~~1 \\ \hline & 4~1 \\ 11 & 2~~~3& \checkmark\\ \hline & 4~3 \\ 12 & 2~~~1 \\ \hline & 1~2 \\ 13 & 3~~~4 \\ \hline & 1~4 \\ 14 & 3~~~2 & \checkmark\\ \hline & 2~1 \\ 15 & 3~~~4 \\ \hline & 2~4 \\ 16 & 3~~~1 & \checkmark\\ \hline & 4~1 \\ 17 & 3~~~2 \\ \hline & 4~2 \\ 18 & 3~~~1 \\ \hline & 1~2 \\ 19 & 4~~~3 \\ \hline & 1~3 \\ 20 & 4~~~2 & \checkmark\\ \hline & 2~1 \\ 21 & 4~~~3 \\ \hline & 2~3 \\ 22 & 4~~~1 & \checkmark\\ \hline & 3~1 \\ 23 & 4~~~2 \\ \hline & 3~2 \\ 24 & 4~~~1 \\ \hline \end{array}\)
The probability is \(\dfrac{8}{24} = \mathbf{\dfrac{1}{3}}\)
