Poisson distribution

1a. The number of fish caught per hour by a particular angler has a Poisson distribution with mean 1.5. Find the probability that he catches exactly 3 fishes in a particular hour.



1b. His son catches 1 fish per hour on average. His also has Poisson distribution and that both of them fish at the same time and their results are independent of one another. Find the probability that
i. In a given hour, they catch a total of at least 2 fishes.
ii. In a 2-hour period, the father catches more fish than his son given that they catch a total of 4 fishes.



1c. Find the length of time, to the nearest minute, for which the probability that they catch at least one fish is more than 0.9.



1d. 50 one-hour periods are chosen at random, and the number of fish caught by the angler in each of the period is recorded. Find the probability that the mean number of fish caught by the angler per hour is more than 2.


What I tried to do is:
1a. X~Po(1.5)
P(X=3) = 0.126

1bi. Y~Po(1)
X+Y~Po(2.5)
P(X+Y ≥2) = 1-P(X+Y ≤1) = 0.713

1bii. 2X~Po(3)
2Y~Po(2)
2X-2Y~Po(1)
P(2X more than 2Y | 2X+2Y=4)
= \(\displaystyle \frac{P(2X-2Y=4)+P(2X-2Y=3)}{1-P(2X+2Y=4)}\)

=0.0778


I don’t know how to do part c and d. But until bii, is what I did correct?

Any help is much appreciated. Thank you.

I'm not quite sure why you picked 2X and 2Y.

Dad:4 Son:0
Dad:3 Son:1

Are the only valid cases.

Catch exactly 4 is Po(2.5,4)

So we have [Po(1.0,0)*P0(1.5,4) + Po(1.0,1)*P0(1.5,3)]/Po(2.5,4)

c. 2.5/60 = 0.4166666... = R = Rate / Minute

Po(R,0) = 0.959
Po(2R,0) = 0.920
Po(4R,0) = 0.846
Po(8R,0) = 0.717
Po(16R,0) = 0.513
Po(32R,0) = 0.264
Po(40R,0) = 0.189
Po(48R,0) = 0.135
Po(52R,0) = 0.115
Po(54R,0) = 0.105
Po(55R,0) = 0.101

I'm going with 56 minutes.

i see.
thank you.