Revision as of 17:05, 20 October 2008 by Kpesyna (Talk)

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The probability of it landing heads the second time, given that it landed heads the first time, is the expected value of the PDF: fQ|H(q). So if you did the problem this way by solving for that PDF, you can solve for the probability by taking the expected value. Expected value remember is:

$ \operatorname{E}(X) = \int_{-\infty}^\infty x f(x)\, \operatorname{d}x . $

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