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Answer

  • P(fish for \(> 2\) hours) = \(P(k=0, \tau=2)\) = \(e^{-0.6 \times 2}\)

  • P(fish for \(> 2\) but \(< 5\) hours) = P(first catch in \([2,5]\) hours) = \(P(k=0,\tau=2)(1-P(k=0,\tau=3)\) which is no fish in \([0,2]\) but at least \(1\) fish in the next \(3\) hours (which will be independent of first \(2\) hours)

  • P(catch at least two fish) = P(at least \(2\) catches before \(2\) hours) = \(1 - P(k=0,\tau=2) - P(k=1,\tau=2)\)

  • \(E\[fish\]\) has two possibilities, either single fish after \(2\) hours, or many fist before \(2\) hours.

\[ \begin{aligned} E\[fish\] &= E\[fish \vert \tau \leq 2\](1-P(\tau > 2)) + E\[fish \vert \tau > 2\] P(\tau > 2)\newline &= (0.6 \times 2) \times (1-P(k=0,\tau=2)) + 1 \times P(k=0,\tau=2) \end{aligned} \]

  • \(E\[\)Total fishing time\(\]\) = \(2 + P(k=0,\tau=2)\frac{1}{\lambda}\), since we fish for atlest \(2\) hours

  • \(E\[\)future fishing time \(\vert\) fished for two hours\(\]\) can be obtained using the memoryless property of Poisson process. The expected time till first arrival is independent of what has happened till now. Thus, \(E[T_{1}] = \frac{1}{\lambda}\)