The occurrence of random events (such as phone calls or e-mail messages) is
often idealized using an exponential distribution. If \(\lambda\) is the average
rate of occurrence of such an event, assumed to be constant over time, then
the average time between occurrences is \(\lambda^{-1}\) (for example, if phone
calls arrive at a rate of \(\lambda=2 /\) min, then the mean time between phone
calls is \(\lambda^{-1}=\frac{1}{2} \mathrm{min}\) ). The exponential
distribution is given by \(f(t)=\lambda e^{-\lambda t},\) for \(0 \leq t<\infty\)
a. Suppose you work at a customer service desk and phone calls arrive at an
average rate of \(\lambda_{1}=0.8 /\) min (meaning the average time between
phone calls is \(1 / 0.8=1.25 \mathrm{min}\) ). The probability that a phone
call arrives during the interval \([0, T]\)
is \(p(T)=\int_{0}^{T} \lambda_{1} e^{-\lambda_{1} t} d t .\) Find the
probability that a phone call arrives during the first 45 s \((0.75\) min) that
you work at the desk.
b. Now suppose that walk-in customers also arrive at your desk at an average
rate of \(\lambda_{2}=0.1 /\) min. The probability that a phone
$$p(T)=\int_{0}^{T} \int_{0}^{T} \lambda_{1} e^{-\lambda_{1} t} \lambda_{2}
e^{-\lambda_{2} x} d t d s$$
Find the probability that a phone call and a customer arrive during the first
45 s that you work at the desk.
c. E-mail messages also arrive at your desk at an average rate of
\(\lambda_{3}=0.05 /\) min. The probability that a phone call and a customer and
an e-mail message arrive during the interval \([0, T]\) is
$$p(T)=\int_{0}^{T} \int_{0}^{T} \int_{0}^{T} \lambda_{1} e^{-\lambda_{1} t}
\lambda_{2} e^{-\lambda_{2} s} \lambda_{3} e^{-\lambda_{3} u} d t d s d u$$
Find the probability that a phone call and a customer and an e-mail message
arrive during the first 45 s that you work at the desk.
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