Suppose to have two meetings in a day. The first meeting is scheduled for $10.00$ a.m., the second one is scheduled for $11.00$ a.m. Let $X_1$ the duration of the first meeting and $X_2$ the duration of the second meeting. $X_1$ and $X_2$ are independent exponential random variables with $E[X_i]=60$(minutes), $i=1,2$.
(a) Determinate the probability of late arrival at the second meeting.
(b) The second meeting begins when the first one is finished. Determine the mean time of second meeting's late (the late is equal to zero if the first meeting'duration is less of $60$ minutes).
To resolve the point (a) I do in this way:
Let $Y=P($late arrival at the second meeting$)$.
$Z=P($get on time at the second meeting$)=P(X_1\leq60)=1-e^{-\lambda x}=1-e^{-1}$
$Y=1-Z=e^{-1}$
It's correct?
For the point (b) I don't know how I can do to resolve. So I hope somebody can help me. Thank you so much.
Be more precise with your task at b, but when I did the right interpretation, at b you search for:
$\Bbb{E}[\Bbb{1}_{\{X_1 > 60\}} (X_1-60)] = \int\limits_{60}^\infty \lambda e^{-\lambda x}(x - 60)dx$
Here $X_1 \sim Exp(\lambda)$ and with your conditions it's easy to see that $\lambda = 60^{-1}$.