Accidents occur in a factory at a rate of 2 per week: $(X \sim Po(2)).$
What is the probability that the time to the first accident is greater than 2 weeks?
What is the probability that the time to the first accident is less than 2 days?
What I've tried so far:
There are 4 accidents every 2 weeks, so $X \sim Po(4)$. I think the question is meaning to ask what the probability is that there will be $0$ accidents in the first 2 weeks, so the poisson rv would be $0$, making the equation —
$$P(X = 0) = \frac{e^{-4}4^0}{0!} = 0.0183$$
for the second one, there are $0.5714$ accidents in $2/7$ of a week (2 days), so $X \sim Po(0.5174)$, but I think the question is asking what the probability of one accident on the first day will be, so that would make $0.2857$ accidents in one day, and the poisson rv would be $1$, so $X \sim Po(0.2857)$, making the equation —
$$P(X = 1) = \frac{e^{-0.2857} 0.2857^1}{1!} = 0.2147$$
Is my thinking correct? should I be looking at these problems from a different perspective?
Best Answer
A Poisson random variable is frequently used to model the probability of observing a count or frequency; i.e., "how many times did something happen?" It is a discrete random variable--we don't say "we observed exactly 3.4 car accidents in three days," or "there were 4.7 bad apples in a bucket."
An exponential random variable is used to model the probability of a failure time, or event time; i.e., "how long did it take for something to happen?" It is a continuous random variable because we can measure time in very fine increments that do not have to be integers; e.g., "the phone failed after 3.26 years of use."
When a Poisson process is used to model the occurrence of an event of interest, it happens that the frequency distribution is Poisson, but the interarrival times are exponentially distributed; that is, the random number of events that occur in a given period of time follows a Poisson distribution, but the random time that elapses between one observed event and the next is exponentially distributed.
Therefore, the time $T$ that elapses until you observe the first factory accident is exponentially distributed with rate parameter $\lambda = 2$ accidents per week. The survival function of $T$ is $S_T(t) = e^{-\lambda t}$, so we want $$\Pr[T > 2] = S_T(2) = e^{-2(2)} = e^{-4}.$$ The other part of your question is addressed similarly, except you need to convert $2$ days into an equivalent (fractional) number of weeks.