Question

A fisherman catches fish according to a Poisson process with rate lambda = 0.6 per hour. The fisherman will keep fishing for two hours. At the end of the second hour, if he has caught at least one fish, he quits; Otherwise, he continues until he catches one fish. (a) Find the probability that he stays for more than two hours. (b) Find the probability that the total time he spends fishing is between two and five hours. (c) Find the expected number offish that he catches. (d) Find the expected total fishing time, given that he has been fishing for four hours.

Answer 1

Answer:

Step-by-step explanation:

Given that a fisherman catches fish according to a Poisson process with rate lambda = 0.6 per hour.

The fisherman will keep fishing for two hours.

Since he continues till he gets atleast one fish, we can calculate probability as follows:

(a) Find the probability that he stays for more than two hours.

= Prob (x=0) in I two hours and P(X≥1) in 3rd hour

=P(x=0)*P(X=0)*P(X≥1) (since each hour is independent of the other)

= $0.5488^2*(1-0.8781)\\\\=0.2645$

(b) Find the probability that the total time he spends fishing is between two and five hours.

Prob that he does not get fish in I two hours * prob he gets fish between 3 and 5 hours

=$P(0)^2 *F(1)^3\\=0.5488^2*0.2645^3\\=0.00557$

(c) Find the expected number offish that he catches.

Expected value in Geometric distribution = $\frac{1-p}{p}$, where p = prob of getting 1 fish in one hour

= $\frac{0.6}{1-0.6} \\=3$

(d) Find the expected total fishing time, given that he has been fishing for four hours.

= Expected fishing time total/expected fishing time for 4 hours

=3/0.6*4

= 1.25 hours