Chapter 5: Problem 5
The lifetime of a radio is exponentially distributed with a mean of ten years. If Jones buys a ten-year-old radio, what is the probability that it will be working after an additional ten years?
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Chapter 5: Problem 5
The lifetime of a radio is exponentially distributed with a mean of ten years. If Jones buys a ten-year-old radio, what is the probability that it will be working after an additional ten years?
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In Example \(5.3\) if server \(i\) serves at an exponential rate \(\lambda_{i}, i=1,2\), show that $$ P\\{\text { Smith is not last }\\}=\left(\frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}}\right)^{2}+\left(\frac{\lambda_{2}}{\lambda_{1}+\lambda_{2}}\right)^{2} $$
A viral linear DNA molecule of length, say, 1 is often known to contain a
certain "marked position," with the exact location of this mark being unknown.
One approach to locating the marked position is to cut the molecule by agents
that break it at points chosen according to a Poisson process with rate
\(\lambda .\) It is then possible to determine the fragment that contains the
marked position. For instance, letting \(m\) denote the location on the line of
the marked position, then if \(L_{1}\) denotes the last Poisson event time
before \(m\) (or 0 if there are no Poisson events in \([0, m])\), and \(R_{1}\)
denotes the first Poisson event time after \(m\) (or 1 if there are no Poisson
events in \([m, 1])\), then it would be learned that the marked position lies
between \(L_{1}\) and \(R_{1}\). Find
(a) \(P\left\\{L_{1}=0\right\\}\),
(b) \(P\left\\{L_{1}
Prove that
(a) \(\max \left(X_{1}, X_{2}\right)=X_{1}+X_{2}-\min \left(X_{1},
X_{2}\right)\) and, in general,
(b) \(\max \left(X_{1}, \ldots, X_{n}\right)=\sum_{1}^{n} X_{i}-\sum_{i
Let \(X_{1}, X_{2}, \ldots, X_{n}\) be independent and identically distributed exponential random variables. Show that the probability that the largest of them is greater than the sum of the others is \(n / 2^{n-1} .\) That is, if $$ M=\max _{j} X_{j} $$ then show $$ P\left\\{M>\sum_{i=1}^{n} X_{i}-M\right\\}=\frac{n}{2^{n-1}} $$ Hint: What is \(P\left\\{X_{1}>\sum_{i=2}^{n} X_{i}\right\\} ?\)
A store opens at 8 A.M. From 8 until 10 customers arrive at a Poisson rate of four an hour. Between 10 and 12 they arrive at a Poisson rate of eight an hour. From 12 to 2 the arrival rate increases steadily from eight per hour at 12 to ten per hour at 2 ; and from 2 to 5 the arrival rate drops steadily from ten per hour at 2 to four per hour at \(5 .\) Determine the probability distribution of the number of customers that enter the store on a given day.
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