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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)

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    $\begingroup$ Hint: To catch exactly $3$ in two hours, the fellow must have caught $0$ in the first and $3$ in the second, or $3$ in the first and $0$ in the second, or.... Just work work out all the cases and add the probabilities. $\endgroup$ – lulu Sep 19 '16 at 23:45
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    $\begingroup$ Alternatively: The number the person catches in $2$ hours is again Poisson, with mean $2\times 1.8$. $\endgroup$ – lulu Sep 19 '16 at 23:47
  • $\begingroup$ @lulu why exactly though do you just multiply the λ by 2? Im not quite understanding the reasoning behind this $\endgroup$ – Nick Arutunov Sep 20 '16 at 0:10
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    $\begingroup$ Then use the first method. But you can read about the sum of Independent Poisson processes here or here $\endgroup$ – lulu Sep 20 '16 at 0:13
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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); the 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.

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