Question

Bob catches an average of 3 fish in 6 hours at his favorite fishing hole. Find the probability that he catches exactly 4 fish in 6 hrs, the probability that he catches at least 2 fish in 6 hours and the probability that he doesn't catch any fish in 4 hrs

Answer 1

Here bob catches an average of 3 fish in 6 hours at his favorite fishing hole.

so here the distribution of catching x number of fishes will be poisson with parameter 6.

p(x) = e^-3 3^x/x! ; x = 0,1,2.....

Pr(x = 4) = e^-3 3⁴/4! = 0.1680

Pr(At least 2 fish) = 1 - Pr(less than 2 fish) = 1 - P(0) - P(1)

= 1 - e^-3 3⁰/0! - e^-3 3¹/1!

= 1 - 0.0498 - 0.1494 = 0.8009

Now expeted number of fish cauth in 4 hrs = 4 * 3/6 = 2

so here if y is the number of fishes caught in 4 hrs then y ~ POISSON(2)

Pr(x = 0) = e^-2 = 0.1353