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

a)

expected number of fish in 30 minutes from 1:50 to 2:30 PM(40 minutes) =40*3/60 =2

P(catch the next fish before 2:30 PM )=P(X>=1) =1-P(X=0)=1-e^-2*2⁰/0! =0.8647

b)

expected fish between 9:00 and 9:30 min (30 min) =30*3/60=1.5

expected fish between 9:00 and 9:50 min (50 min) =50*3/60=2.5

P(2nd fish between 9:30 and 9:50) =P(2 or more fish till 9:50 min) -P(2 or more fish till 9:30 min)

=(1-e^-2.5*2.5⁰/0!-e^-2.5*2.5¹/1!)-(1-e^-1.5*1.5⁰/0!-e^-1.5*1.5¹/1!)

=0.2705