I think this is an erlang distribution problem but i'm not sure.

A fisher expects to catch a fish every 25 minutes.

What is the probability that she will need to wait 2 hours to catch 4 fish?

What is the probability that she will need to wait between 3 and 5 hours to catch 8 fish?

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    $\begingroup$ Formatting tips here. $\endgroup$ – Em. Feb 2 '16 at 3:01

I believe you can model the waiting time of a fish (catching a fish) as an exponential distribution with mean $25$ minutes. I think you would know that the sum of independent exponential waiting times follows a gamma (Erlang) distribution $T$.

Then let $T_i$ denote the $i$th waiting time. The problems become

  1. $P(T_4>2\text{ hrs})$

  2. $P(3\text{ hrs}<T_8<5 \text{ hrs})$

  • $\begingroup$ So i use the CDF for erlang: $$1- (e^(-k \theta x)(k(\theta)x)i)/i! $$ ? $\endgroup$ – Mike Feb 2 '16 at 2:58
  • $\begingroup$ @Mike I believe, you did not copy that correctly, but yes, you could use the cdf. That is one approach for part 1. For part 2, I would use the differenece of cdfs. $\endgroup$ – Em. Feb 2 '16 at 3:03
  • $\begingroup$ ok how do I calculate the CDF? for the first one do I use k=4 $\theta$= 25 x=120 and calculate for i = 0,1, and 2 then add them together? Sorry I haven't done probabilities in a while. Im a computer science major and need them for simulation... $\endgroup$ – Mike Feb 2 '16 at 3:06
  • $\begingroup$ @Mike Yes, the typical formula for the cdf is $$1-e^{-\lambda x}\sum_{k=0}^{r-1}\frac{(\lambda x)^k}{k!}$$ where $r$ is the number of arrivals and $\lambda$ is the rate. So there is some adding up to do. $\endgroup$ – Em. Feb 2 '16 at 3:10
  • $\begingroup$ This is the formula I have: $$1-\sum_{i=0}^{k-1}\frac{e^(-k\theta x)*(k\theta x)^i)}{i!}$$ $\endgroup$ – Mike Feb 2 '16 at 3:16

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