If the number of fish a person catches per hour at Woods Canyon Lake is a random variable having the Poisson distribution with λ=1.8, find the probability that a person fishing there will catch 3 fish in 2 hours. (Answer is .2125, but I have no idea how to do it)
In applications of the Poisson distribution, it is important to keep an eye on the 'domain' in space and/or time for which the rate $\lambda_1$ is specified.
In your problem $\lambda_1$ is in terms of fish per hour. So if your question involves two hours, you need to adjust it to fish per 2 hours: $\lambda_2 = 2\lambda_1.$ Then $P(X_2 = 3) = 0.2125.$ Computation in R software below:
lam.1 = 1.8; lam.2 = 2*lam.1 dpois(3, lam.2) ## 0.2124693
And if you were asked about how many fish are caught in an eight-hour period of steady fishing, it would be $\lambda_8 = 8\lambda_1.$
The other method suggested by @lulu is correct, but it is important for you to understand this method of adjusting the rate to match the problem. Many textbook and real-life applications require this crucial step. (And if you later explore the connection between the Poisson and exponential distributions, this adjustment is crucial to the argument.)
I don't know if you are familiar with simulations. But here is a simulation
that uses the 'addition' method. I simulate a million first hours (v's) and
a million second hours (w's), both using $\lambda_1 = 1.8.$ Then check to
what proportion of the time $U + V$ turns out to be 3. The vector
v + w == 3
has either the logical value TRUE (if the sum is 3) or FALSE (if not);
mean of a logical vector its its proportion of TRUE's.
v = rpois(10^6, 1.8); w = rpois(10^6, 1.8) mean(v + w == 3) ## 0.212628
Confession: As it happened, this simulation turned out to be accurate to almost four places. Most simulations with a million iterations would have been a little less accurate. The 95% margin of simulation error is 0.0025.