(b) If we encounter 1 rabbit every 1.5 hours, then that means we'll encounter 0.67 rabbits every hour (pretty gory). We need to integrate e^{- x} = 0.67 e^{-0.67 x} between hours 7 and 8. This gives 0.0045 (unlikely, since we'll probably encounter the first one much earlier in the day). The exponential distribution applies to this problem, since we are waiting for the first event to occur when events happen randomly and continuously over time.

(c) P = 0.999⁵⁰⁰ = 0.606 for zero errors. P = 500*0.999⁴⁹⁹*0.001 = 0.303 for one error. P = 1-0.606-0.303 = 0.091 for more than one error. This problem describes a binomial distribution. Alternatively, since the probability of an error is small, you could use a Poisson distribution to answer this problem.

(d) E[X] = 10/0.1 = 100 minutes since = 0.1 fish per minute. Var(X) = r/ ² = 10/0.01 = 1000 so the standard deviation is 31.6 minutes. From the normal approximation, the 95% confidence limits around the mean of 100 minutes is 100-2*31.6 to 100+2*31.6, or roughly 40 to 160 minutes until you go home.