Choose A and B. Then choose P and Q at random, then draw lines AP and BQ. This is equivalent to choosing a random direction (since the union of all rays leading from a point is the whole plane, and every ray has the same number of points, uncountably infinite.)
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As the problem stated, disregard the functionally impossible case that AP and BQ are parallel, or that their intersection, X, equals any of the other four points.
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Place points P' and Q' on AP and BQ respectively, such that A is between P and P' and B is between Q and Q'. Now rays AP and AP' are opposites and each has a 1/2 probability of being chosen, and exactly one of them contains X. Rays BQ and BQ' also each have a 1/2 probability, and exactly one of those contains X. If both rays that contain X are chosen, they  intersect, but if not, they do not. The probability of this happening is (1/2)*(1/2) = 1/4.