Revision as of 09:21, 4 May 2023 by Admin (Created page with "'''Solution: B''' Let X be the number of burglaries. Then, <math display = "block"> \begin{align*} \operatorname{E}(X | X \geq 2) = \frac{\sum_{x=2}^{\infty}xp(x)}{1-p(0)-p(...")
Exercise
ABy Admin
May 04'23
Answer
Solution: B
Let X be the number of burglaries. Then,
[[math]]
\begin{align*}
\operatorname{E}(X | X \geq 2) = \frac{\sum_{x=2}^{\infty}xp(x)}{1-p(0)-p(1)} &= \frac{\sum_{x=0}^{\infty}xp(x)-p(0)-(1)p(1)}{1-p(0)-p(1)} \\
&= \frac{1-p(1)}{1-p(0)-p(1)} \\
&= \frac{1-e^{-1}}{1-e^{-1}-e^{-1}} = 2.39.
\end{align*}
[[/math]]