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2017 Paper 2 Q12
D: 1600.0 B: 1563.6
probability Poisson distribution binomial distribution sum of independent variables conditional distribution expected value exponential distribution minimum and maximum of random variables

Adam and Eve are catching fish. The number of fish, \(X\), that Adam catches in any time interval is Poisson distributed with parameter \(\lambda t\), where \(\lambda\) is a constant and \(t\) is the length of the time interval. The number of fish, \(Y\), that Eve catches in any time interval is Poisson distributed with parameter \(\mu t\), where \(\mu\) is a constant and \(t\) is the length of the time interval The two Poisson variables are independent. You may assume that the expected time between Adam catching a fish and Adam catching his next fish is \(\lambda^{-1}\), and similarly for Eve.

  1. By considering \(\P( X + Y = r)\), show that the total number of fish caught by Adam and Eve in time \(T\) also has a Poisson distribution.
  2. Given that Adam and Eve catch a total of \(k\) fish in time \(T\), where \(k\) is fixed, show that the number caught by Adam has a binomial distribution.
  3. Given that Adam and Eve start fishing at the same time, find the probability that the first fish is caught by Adam.
  4. Find the expected time from the moment Adam and Eve start fishing until they have each caught at least one fish.
[Note This question has been redrafted to make the meaning clearer.]

Solution:

  1. \(\,\) \begin{align*} && \mathbb{P}(X+Y=r) &= \sum_{k=0}^r \mathbb{P}(X = k, Y = r-k) \\ &&&= \sum_{k=0}^r \mathbb{P}(X = k)\mathbb{P}( Y = r-k) \\ &&&= \sum_{k=0}^r \frac{e^{-\lambda T} (\lambda T)^k}{k!}\frac{e^{-\mu T} (\mu T)^{r-k}}{(r-k)!}\\ &&&= \frac{e^{-(\mu+\lambda)T}}{r!}\sum_{k=0}^r \binom{r}{k}(\lambda T)^k (\mu T)^{r-k}\\ &&&= \frac{e^{-(\mu+\lambda)T}((\mu+\lambda)T)^r}{r!} \end{align*} Therefore \(X+Y \sim Po \left ( (\mu+\lambda)T \right)\)
  2. \(\,\) \begin{align*} && \mathbb{P}(X = r | X+Y = k) &= \frac{\mathbb{P}(X=r, Y = k-r)}{\mathbb{P}(X+Y=k)} \\ &&&= \frac{\frac{e^{-\lambda T} (\lambda T)^r}{r!}\frac{e^{-\mu T} (\mu T)^{k-r}}{(k-r)!}}{\frac{e^{-(\mu+\lambda)T}((\mu+\lambda)T)^k}{k!}} \\ &&&= \binom{k}{r} \left ( \frac{\lambda}{\lambda + \mu} \right)^r \left ( \frac{\mu}{\lambda + \mu} \right)^{k-r} \end{align*} Therefore \(X|X+Y=k \sim B(k, \frac{\lambda}{\lambda + \mu})\)
  3. \(P(X=1|X+Y = 1) = \frac{\lambda}{\lambda + \mu}\)
  4. Let \(X_1, Y_1\) be the time to the first fish are caught by Adam and Eve, then \begin{align*} && \mathbb{P}(X_1, Y_1 > t) &= \mathbb{P}(X_1> t) \mathbb{P}( Y_1 > t) \\ &&&= e^{-\lambda t}e^{-\mu t} \\ &&&= e^{-(\lambda+\mu)t} \\ \Rightarrow && f_{\max(X_1,Y_1)}(t) &= (\lambda+\mu)e^{-(\lambda+\mu)} \end{align*} Therefore the expected time is \(\frac1{\mu+\lambda}\)

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