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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Let alpha=angle(OAB), then OB=2*sen(alpha/2).
ReplyDeleteVariable 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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