1. Quote of the week: "A good rule to remember for life is that when it comes to plastic surgery and sushi, never be attracted by a bargain" - Graham Norton
2. New quiz AQQ263 about probability puzzle on a square (this one is a complex one...)
Good luck!
P.S. Please spread those quizzes among colleagues, friends or family members; we want more participants!
Probability was never my thing but here’s my try at the problem.
Each point (A or B) are independent events so they have a 0.25 chance of landing on one of the four sides. This creates 16 states, each with the same chance of (0.25)x(0.25)=0.0625. If they land on the same side (4 states), the distance between them cannot be >C so they don’t contribute to the result. If they land on opposite sides (4 states) the distance between them can be >= C so they add ~0.25 to the result. The last group of 8 land on adjacent sides so the distance between them is contingent on where they locate and this is the hard part of the problem. Refer to the problem figure. The points form a right triangle as shown. Fix length d at C so anything outside of it forms a line >C. The probability of b outside the triangle is dependent on x, the length of the triangle base (0 to a).
I reviewed my analysis and found an error in my evaluation of the integral. It should be 0.2146, not 0.7854. That said, my answer is 0.3573, which agrees with the above. 
