Poisson and Cumulative Distribution Double Check this is simply me double checking my answer.
Let Y be number of fish caught on a trip with Poisson distribution and $\lambda=4$.
What is prob of catching 3 or fewer trout on trip?
I said this was $\Sigma^3_{i=0}e^{-\lambda}*\frac{\lambda^i}{i!}=e^{-4}+4e^{-4}+8e^{-4}+\frac{4^3e^{-4}}{6}$. Correct?
What is the prob that a fish's length is greater than 9 inches where the length is distributed between 6 and 12 with density $f(x)=\frac{12}{x^2}$.
I said this was $P(X>9)=1-P(X\le9)=1-\int^9_6f(x)=1-\frac{2}{3}=1/3.$ Correct?
Lastly, given that on the trip we caught 3 or fewer trout, what is the prob that none were keepers. They are a keeper if length is greater than 9 inches.
These events are independent. So we have $P(X\le9|Y\le3)=\frac{P(X\le9,Y\le3)}{P(Y\le3)}=\frac{P(X\le9)P(Y\le3)}{P(Y\le3)}=P(X\le9)=2/3$ because it is the integral from $6\rightarrow9$ of $f(x)$. Correct?
 A: You have the first two parts right, and some difficulties on the third.
Here is an outline of how to put it all together for correct answers
all the way through. You should verify the numerical results.
(1) $P(\mathrm{3\; or\; fewer\; caught}) =  0.1953668,$ following your Poisson formula.
(2) $P(\mathrm{keeper}) = P(X \ge 9) = 1/3,$ by your integration. 
Note that the CDF is $F(x) = 2 - 12/x,$ for $6 < x < 12.$
In particular, $F(6) = 0$ and $F(12) = 1,$ as required of a CDF.
(3) The probability that three or fewer fish were caught and none were keepers is 
$$\sum_{k=0}^3 P\{Y = k\}P\{k \mathrm{\;underweight}\} 
= \sum_{k=0}^3 \frac{e^{-4}4^k}{4!}p^k = 0.1528430,$$
where $p = P\{X < 9\} = 2/3,$ as in the comment by @Did. This is
due to the Law of Total Probability and the independence of $X$ and $Y$.
So $P(\mathrm{No\; keepers}|\mathrm{Three\; or\; fewer\; caught}) = 0.1528430/0.1953668 \approx 0.7823,$
by the definition of conditional probability. Notice that the probability of having 'no keepers' depends on how many fish are caught.
