Question

In a lake, Peter catches fishes in an average rate of 3 fishes per hour. (Assume Peter always catches one fish at a time and the events of catching each fish are independent.)

(a) [10 points] At 2:00PM, if it’s known that Peter caught the previous fish at 1:50PM, what is the probability that he will catch the next fish before 2:30PM?

(b) [10 points] Assume Peter starts catching fish at 9:00AM, what is the probability that he will catch the 2nd fish between 9:30 and 9:50AM?

Answer 1

We are getting the fish catching rate here as:

a)Poisson distribution follows a memoryless property, therefore given that Peter caught the previous fish at 1:50PM, it does change the future probabilities here. The probability that he will catch the next fish before 230 is computed as:

= 1 - Probability that there is not fish caught in the next 30 min = 0.5 hours

= 1 - e^-3*0.5 = 1 - e^-1.5 = 0.7769
Therefore 0.7769 is the required probability here.

b) Probability he will catch the second fish between 930 and 950 am is computed here as:

= Probability that one fish is caught between 9am and 930am * (1 - Probability that no fish is caught in the next 20 minutes)

= 1.5e^-1.5*(1 - e^-3/3)

= 1.5e^-1.5*(1 - e^-1) = 0.2116

Therefore 0.2116 is the required probability here.