Poisson Distribution

Showing 26-26 of 26 problems
1989 Paper 1 Q15
D: 1500.0 B: 1516.0

I can choose one of three routes to cycle to school. Via Angle Avenue the distance is 5\(\,\)km, and I am held up at a level crossing for \(A\) minutes, where \(A\) is a continuous random variable uniformly distributed between \(0\) and 10. Via Bend Boulevard the distance is 4\(\,\)km, and I am delayed, by talking to each of \(B\) friends for 3\(\,\)minutes, for a total of \(3B\) minutes, where \(B\) is a random variable whose distribution is Poisson with mean 4. Via Detour Drive the distance should be only 2\(\,\)km, but in addition, due to never-ending road works, there are five places at each of which, with probability \(\frac{4}{5},\) I have to make a detour that increases the distance by 1\(\,\)km. Except when delayed by talking to friends or at the level crossing, I cycle at a steady 12\(\,\)km\(\,\)h\(^{-1}\). For each of the three routs, calculate the probability that a journey lasts at least 27 minutes. Each day I choose one of the three routes at random, and I am equally likely to choose any of the three alternatives. One day I arrive at school after a journey of at least 27 minutes. What is the probability that I came via Bend Boulevard? Which route should I use all the time: \begin{questionparts} \item if I wish my average journey time to be as small as possible; \item if I wish my journey time to be less than 32 minutes as often as possible? \end{questionpart} Justify your answers.

Show Solution
\(A \sim 5\cdot 5 + U[0,10]\) \(B \sim 4 \cdot 5 + 3 \textrm{Po}(4)\) \(C \sim 2 \cdot 5 + B(5, \frac{4}{5}) \cdot 5\) \begin{align*} && \mathbb{P}(A \leq 27) &= \mathbb{P}(U \leq 2) = 0.2 \\ && \mathbb{P}(B \leq 27) &= \mathbb{P}(3 \textrm{Po}(4) \leq 7) \\\ &&&= \mathbb{P}(Po(4) \leq 2) \\ &&&= e^{-4}(1 + 4 + \frac{4^2}{2}) \\ &&&= 0.23810\ldots \\ && \mathbb{P}(C \leq 27) &= \mathbb{P}(5 \cdot B(5,\tfrac45) \leq 17) \\ &&&= \mathbb{P}(B(5,\tfrac45) \leq 3) \\ &&&= \binom{5}{0} (\tfrac15)^5 + \binom{5}{1} (\tfrac45)(\tfrac 15)^4+ \binom{5}{2} (\tfrac45)^2(\tfrac 15)^3 + \binom{5}3 (\tfrac45)^3(\tfrac 15)^2+\\ &&&= 0.26272 \end{align*} \begin{align*} \mathbb{P}(\text{came via B} | \text{at least 27 minutes}) &= \frac{\mathbb{P}(\text{came via B and at least 27 minutes})}{\mathbb{P}(\text{at least 27 minutes})} \\ &= \frac{\frac13 \cdot 0.23810\ldots }{\frac13 \cdot 0.2 + \frac13 \cdot 0.23810\ldots + \frac13 \cdot 0.26272} \\ &= 0.3397\ldots \\ &= 0.340 \, \, (3\text{ s.f.}) \end{align*}
  1. \begin{align*} \mathbb{E}(A) &= 25 + 5 &= 30 \\ \mathbb{E}(B) &= 20 + 3\cdot 4 &= 32 \\ \mathbb{E}(C) &= 10 + 5 \cdot 4 &= 30 \end{align*} \(A\) and \(C\) are equally good.
  2. \begin{align*} \mathbb{P}(A \leq 32) &= \mathbb{P}(U \leq 7) &= 0.7 \\ \mathbb{P}(B \leq 32) &= \mathbb{P}(Po(4) \leq 4) \\ &= e^{-4}(1 + 4 + 8 + \frac{4^3}{6}) &= 0.4334\ldots \\ \mathbb{P}(C \leq 32) &= \mathbb{P}(B(5,\tfrac45) \leq 4) \\ &= 1 - \mathbb{P}(B(5,\tfrac45) = 5) \\ &= 1 - \frac{4^5}{5^5} &=0.67232 \end{align*} So you should choose route \(A\).