Question

Consider a point on the Trans-Australian Highway, where two old wombats live. Arrivals of cars at this point follow a Poisson distribution; the average rate of arrivals is 1 car per 12 seconds. 1. One of these old wombats requires 12 seconds to cross the highway, and he starts out immediately after a car goes by. What is the probability he will survive?
(A) 0.114
(B) 0,736
(C) 0.368
(D) 0.184

Answer 1

Answer:

(C) 0.368

Step-by-step explanation:

The Poisson distribution models the probability that an event occurs k times in a given fix interval of time.

Is defined by the probability function

$\huge P(k)=\frac{\lambda^ke^{-\lambda}}{k!}$

where

k = number of events in the given interval

$\huge \lambda$ = average number of events in the given interval.

In our specific case, the given interval of time is 12 seconds, and $\huge \lambda$ = 1 (1 car  every 12 seconds).

As the old wombat requires 12 seconds to cross the highway, it might survive if only 1 car arrives over those 12 seconds.

The probability of this fact is P(1)

$\huge P(1)=\frac{1^1e^{-1}}{1!}=e^{-1}=0.367879\approx 0.368$

So, the right answer is (C), 0.368