Question

Jack likes to go fishing. While waiting for the fishes to bite, he formulates the following model for the process: fishes bite according to a Poisson process with intensity 4 bites per hour. Biting fishes are caught independently, and on average only one in two times.
a) What is the probability that six fishes bite during the first two hours?
b) What is the probability that he fails to catch any fishes during the first two hours?
c) What is the probability that, during the first two hours, six fishes bite and two of these are caught?

Answer 1

Answer:

a) 0.122

b) 3.35 × 10⁻⁴

c) 0.1071

Step-by-step explanation:

Here we have the Poisson process given by

$P(N(t) = k) = \frac{e^{-\lambda t} (\lambda t)^k }{k!}$

Where:

λ = Rate = 4 bites/hour and

t = Length of time that is

Mean = λt

Where k = 6 fishes bites and t = 2 hours we have

$P(N(2) = 6) = \frac{e^{-4 \times 2} (4 \times 2)^6 }{6!} = 0.122$

b) The probability that he fails to catch any fishes during the first two hours is the probability t that he catches 0 fish as follows

$P(N(2) = 0) = \frac{e^{-4 \times 2} (4 \times 2)^0 }{0!} = 3.35 \times 10^{-4}$

(c) Here we have

$P(N(2) = 6) = 0.122$

For 1 fish caught in two times we have first time no fish and second time he caches a fish gives

0.122×(1-0.122) = 0.1071