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A fisherman catches fish of type $A$ and $B,$ determined by Poisson processes of rythms $a$ and $b$ / minute, respectively.

(1) If the fisherman caught $10$ fish in $2$ hours, what is the probability he caught none between $30$ min and $90$ min?

(2) If the fisherman caught $10$ fish in $2$ hours, what is the probability he caught none of type $A$ between $30$ min and $90$ min?

(3) If $T$ is the first time that at least two fish of type $A$ and at least two fish of type $B$ are caught, what is the distribution of the random variable $T$?

Attempt

(1) If $X_t,~Y_t,N_t=X_t+Y_t$ are the number of fish of type A, type B, total, caught by the fisherman, respectively, on the interval $(0,t]$, then we seek $P(N_s-N_r=0~|~N_t=10),$ where $r=30, ~s=90,~ t=120.$ Then $$P(N_s-N_r=0~|~N_t=10)=\frac{P(N_s-N_r=0,~N_t=10)}{P(N_t=10)}$$ and this is where I am stuck since I don't know how to use the independent increments of the Poisson process $\{N_t\}$ - if I had $N_{90}-N_{3},~N_{30}$ things would be easier.

(2) I think solving $(1)$, would help!

Thanks a lot in advance!

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  • $\begingroup$ In your probability you have$[{N_r} = {N_s}$ but I think that you want ${N_r} = {N_s} = 0$. Is that the case? $\endgroup$ – user328032 Apr 21 '16 at 10:26
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1) The conditional distribution of the number of fish caught in $[30,90]$ is $Bin(\frac{1}{2}, 10)$. Because $\frac{90-30}{120-0} = \frac{1}{2}$ and $n=10$. So the wanted probabilty is that all the $10$ fish were caught in $[0,30)\cup(90,120]$ is $\frac{1}{2^{10}}$.

2) Let $N^A(2)$ be the number of type $A$ fish in $[0,120]$. Let $X^A(1)$ be the number of fish in $[30, 90]$. So, \begin{align} P(X^A(1) = 0) =& \sum_{x=0}^{10} P(X^A(1)=0|N^A(2)=x)P(N^A(2)=x)\\ =& \sum_{x=0}^{10} \frac{1}{2^x}\binom{10}{x}\left( \frac{a}{a+b} \right)^x\left( \frac{b}{a+b} \right)^{10-x} \, . \end{align}

3) Let $T_2^A$ be the time when two fish of type $A$ had been caught, so $T_2^A \sim Erlang(2, a) $, similarly $T_2^B \sim Erlang(2, b) $. So, $T = (T_2^A, T_2^B)$ \begin{align} P_T(t,t) = P(T_2^B=t)P(T_2^B=t) = a^2e^{-a t}tb^2e^{-b t}t = (ab)^2e^{-t(a+b)}t^2, \quad t\ge 0 \end{align}

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  • $\begingroup$ thank you! I see. How about the second question? The only thing I know is that, given $N_t=10$ the number of fish of type $A$ caucht is distributed according to the binomial distribution $Bin(10,\frac{a}{a+b}).$ $\endgroup$ – Nikolaos Skout Apr 21 '16 at 12:58
  • $\begingroup$ I've edited the answer $\endgroup$ – V. Vancak Apr 21 '16 at 19:03
  • $\begingroup$ @V.Vancak could you please explain the first part again? $\endgroup$ – user585380 Sep 9 '18 at 15:02
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    $\begingroup$ @user585380 If the fisherman caught $10$ fish in $[0,120]$ and you are interested in the event that in the subinetrval $[30,90]$ he caught none, then in $[0,30)$ and $(90, 120]$ he caught his $10$ fish. Note that $[30,90]$ is exactly half of the whole interval $[0, 120]$, as such if the timing is uniform on $[0,120]$ the probability that it falls out of this interval is $1/2$, the $10$ events are independent, hence $\binom{10}{10} \frac{1}{2^{10}} \cdot \frac{1}{2^0}$. $\endgroup$ – V. Vancak Sep 10 '18 at 8:25
  • $\begingroup$ @V.Vancak thank you so much. Got it now. $\endgroup$ – user585380 Sep 10 '18 at 8:52

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