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Probability - QuestionCove
OpenStudy (darkprince14):

Probability

2 years ago
OpenStudy (darkprince14):

2 years ago
OpenStudy (darkprince14):

@Ashleyisakitty

2 years ago
OpenStudy (anonymous):

\[ E[h(X)] = \sum_{x}h(x)f(x) \]In our case: \[ E[X^{-1}] = \sum_{x=1}^{\infty }\frac{(1-p)^{x-1}p}{x} \]

2 years ago
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