Poisson Probability Question -Please help

Question

Can you please help me on this question The number of fishes caught by a fisherman is Poisson with an average of $6$ fishes per hour, assume he caught $8$ fishes in the last hour, what is the probability of no fish was caught in the first $10$ minutes? Thanks much for your help!

Answer 1

First we need to find the mean number of fishes caught in $10$ minutes. We can find this by converting the mean from hour to minutes: $\mu= \dfrac{6 \text{ fishes}}{1\text{hr}}\times \dfrac{1\text{ hr}}{6 \text{ (10-minutes)}}= \dfrac{1 \text{ fish}}{10 \text{ minutes}}= 1$. Thus we have: $P(x = k) = \dfrac{\mu^k\cdot e^{-\mu}}{k!}$. Plug $k = 0, \mu = 1$ into the formula we have: $P(x = 0) = e^{-1}= 0.368$

Answer 2

Hint: One of the properties* of the Poisson distribution is that under the condition that $n$ events have occurred in the the interval $(0,t)$, the times at which the events occurred are distributed independently and uniformly in $(0,t)$.

So suppose $x_1, x_2, x_3, \dots , x_8$ are chosen independently and uniformly in $(0,1)$. What is the probability that none of them fall in the interval $(0, 1/6)$?

* Reference: Theorem 5.2, Introduction to Probability Models, 9th Edition by Sheldon M. Ross.

Answer 3

Let $A$ be the event that no fish was caught in the first 10 minutes of the hour, and $B$ be the event that 8 fish were caught in the hour.

Since the fish are caught at a rate of 6 per hour on average (equivalently 1 per 10 minutes or 5 per 50 minutes), using Bayes' theorem yields:

$$\mathbb{P}(A|B) = \mathbb{P}(B|A) \cdot \frac{\mathbb{P}(A)}{\mathbb{P}(B)} = \frac{5^8}{8!}e^{-5} \cdot \frac{\frac{1}{0!}e^{-1}}{\frac{6^8}{8!}e^{-6}} = \left(\frac{5}{6}\right)^8 = 0.232568 $$

because $\mathbb{P}(B|A)$ corresponds to the probability of catching 8 fish in 50 minutes. Note that the solution of $(5/6)^8$ can be arrived at elegantly using awkward's answer.