Wednesday, December 2, 2009

Fox 199


  1. alpha=angle(OAB) is uniformly distributed on [0,2pi). Then p=1/(2pi) is the probability of find alpha in that interval. Area of triangle OAB is equal to sen(alpha)/2. The expected value is then the integral of sen(alpha)/(2pi) on [0,pi], that equals to 1/pi.

    Answer is D.

  2. Excel simulation gives the same answer:

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