17 problems found
During his performance a trapeze artist is supported by two identical ropes, either of which can bear his weight. Each rope is such that the time, in hours of performance, before it fails is exponentially distributed, independently of the other, with probability density function \(\lambda\exp(-\lambda t)\) for \(t\geqslant0\) (and 0 for \(t < 0\)), for some \(\lambda > 0.\) A particular rope has already been in use for \(t_{0}\) hours of performance. Find the distribution for the length of time the artist can continue to use it before it fails. Interpret and comment upon your result. Before going on tour the artist insists that the management purchase two new ropes of the above type. Show that the probability density function of the time until both ropes fail is \[ \mathrm{f}(t)=\begin{cases} 2\lambda\mathrm{e}^{-\lambda t}(1-\mathrm{e}^{-\lambda t}) & \text{ if }t\geqslant0,\\ 0 & \text{ otherwise.} \end{cases} \] If each performance lasts for \(h\) hours, find the probability that both ropes fail during the \(n\)th performance. Show that the probability that both ropes fail during the same performance is \(\tanh(\lambda h/2)\).
Solution: This is the memoryless property of the exponential distribution so it has the same distribution as when \(t = 0\). Let \(T\) be the time the rope fails, then \begin{align*} && \mathbb{P}(T > t | T > t_0) &= \frac{\mathbb{P}(T > t)}{\mathbb{P}(T > t_0)} \\ &&&= \frac{e^{-\lambda t}}{e^{-\lambda t_0}} \\ &&&= e^{-\lambda(t-t_0)} \end{align*} This means that each rope (as long as it hasn't broken) can be considered "as good as new". Suppose \(T_1, T_2 \sim Exp(\lambda)\) are the time to failures for each rope, then \begin{align*} && \mathbb{P}(\max(T_1, T_2) < t) &= \mathbb{P}(T_1 < t, T_2 < t) \\ &&&= (1-e^{-\lambda t})^2 \\ \Rightarrow && f(t) &= 2(1-e^{-\lambda t}) \cdot (\lambda e^{-\lambda t}) \\ &&&= 2\lambda e^{-\lambda t}(1-e^{-\lambda t}) \end{align*} Therefore \(\max(T_1, T_2) \sim Exp(2\lambda)\) and the pdf is as described. \begin{align*} && \mathbb{P}(\text{both fail during the }n\text{th}) &= \left ( \int_{(n-1)h}^{nh} \lambda e^{-\lambda t} \d t \right)^2 \\ &&&=\left (\left [ -e^{-\lambda t}\right]_{(n-1)h}^{nh} \right)^2 \\ &&&= \left ( e^{-\lambda (n-1)h}( 1-e^{-\lambda h}) \right)^2 \\ &&&= e^{-2(n-1)h\lambda}(e^{-\lambda h}-1)^2 \\ \\ && \mathbb{P}(\text{both fail in same performance}) &= \sum_{n=1}^{\infty} \mathbb{P}(\text{both fail during the }n\text{th}) \\ &&&= \sum_{n=1}^{\infty}e^{-2(n-1)h\lambda}(e^{-\lambda h}-1)^2 \\ &&&= (e^{-\lambda h}-1)^2 \frac{1}{1-e^{-2h\lambda}} \\ &&&= \frac{e^{-\lambda h}-1}{1+e^{-h\lambda}} \\ &&&= \tanh(\lambda h/2) \end{align*}
When he sets out on a drive Mr Toad selects a speed \(V\) kilometres per minute where \(V\) is a random variable with probability density \[ \alpha v^{-2}\mathrm{e}^{-\alpha v^{-1}} \] and \(\alpha\) is a strictly positive constant. He then drives at constant speed, regardless of other drivers, road conditions and the Highway Code. The traffic lights at the Wild Wood cross-roads change from red to green when Mr Toad is exactly 1 kilometre away in his journey towards them. If the traffic light is green for \(g\) minutes, then red for \(r\) minutes, then green for \(g\) minutes, and so on, show that the probability that he passes them after \(n(g+r)\) minutes but before \(n(g+r)+g\) minutes, where \(n\) is a positive integer, is \[ \mathrm{e}^{-\alpha n(g+r)}-\mathrm{e}^{-\alpha\left(n(g+r)\right)+g}. \] Find the probability \(\mathrm{P}(\alpha)\) that he passes the traffic lights when they are green. Show that \(\mathrm{P}(\alpha)\rightarrow1\) as \(\alpha\rightarrow\infty\) and, by noting that \((\mathrm{e}^{x}-1)/x\rightarrow1\) as \(x\rightarrow0\), or otherwise, show that \[ \mathrm{P}(\alpha)\rightarrow\frac{g}{r+g}\quad\mbox{ as }\alpha\rightarrow0. \] {[}NB: the traffic light show only green and red - not amber.{]}