1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Formatting tips here. $\endgroup$ – Em. Feb 2 '16 at 3:01
0
$\begingroup$

Clarification/Hint:
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. $P(3\text{ hrs}<T_8<5 \text{ hrs})$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.