Saturday, December 5, 2009

Fox 201


  1. Let alpha=angle(OAB), then OB=2*sen(alpha/2).
    Variable alpha is uniformly distributed on [0,2pi]. Then probability of alpha is equals to 1/2pi on that interval.
    The expected value is:
    E[OB]=2*int_0_pi{(1/2pi)*2*sen(alpha/2)} = 4/pi.

    Answer is A.

    Note: int_0_pi = integral between 0 and pi limits.

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