Discrete Probability A fisherman catches fish at a Poisson r

Discrete Probability

A fisherman catches fish at a Poisson rate of two per hour from a large lake with lots of fish. If he starts fishing at 10 AM, what is the probability that he catches one fish by 10.30 AM and three fish by 12 PM?

Solution

As we are given that fisghes catches at poisson rate (2 fishes in 1 hour).

Thus probability of caching n fishes at time t [p(N(t) =n)] = [{ e^(-lamda * t) } *(lamda *t)^n] / n!

where lamda = 2

probability that the 1 fish catches by (1/2) hour is p(N(1/2) = 1]

p(N(1/2) = 1] = [{ e^(-2 * 1/2) } *(2 *1/2)^1] / 1!

= e^(-1)

=0.367879

probability that the 3 fish catches by (2) hours is p(N(2) = 3]

p(N(2) = 3] = [{ e^(-2 * 2) } *(2 *2)^3] / 3!

= [{ e^(-4 } *(4)^3] / (3*2)

= 0.195367