Hi Math Stack Exchange Community,
I am stuck with this question and not sure how can I solve it. Any help would be much appreciated.
You work as part of the maintenance team in a manufacturing company and keep a record of the machine breakdowns. From your past experience, it seems that there are on average 2.5 serious breakdowns every year in your company. Then,
a) The probability that there will be 5 breakdowns in a given year is?
b) The probability that there will be fewer than 4 breakdowns in a given year is?
c) The probability that there will be at least 2 breakdowns in a given year is?
Collecting comments:
A random variable which follows a Poisson distribution takes non-negative integer values and satisfies:
$$\Pr(X=k) = \dfrac{e^{-\lambda}\lambda^k}{k!}$$
where $\lambda$ is the expected value of $X$ and $k$ is the actual number of occurrences we see in the timeframe.
The problem statement here tells us that the average number of breakdowns per year is $2.5$... in other words it tells us that $\lambda = 2.5$ here.
The first problem asks us what the probability is that we see $5$ breakdowns. That is, it asks us for $\Pr(X=5)$ given that $\lambda = 2.5$. That is, as you correctly found in the comments after coaxing:
$$\Pr(X=5)=\dfrac{e^{-2.5}(2.5)^5}{5!}$$
For the second problem, it asks us the probability is we see fewer than $4$ breakdowns. That is to say, that we saw $0,1,2$ or $3$ breakdowns. We note that $\Pr(X< 4) = \Pr(X=0)+\Pr(X=1)+\Pr(X=2)+\Pr(X=3)$, use the values of $0,1,2,3$ in place of $k$ for each of the respective calculations, all still using the lambda value of $\lambda = 2.5$, and conclude.
For the final problem, it asks us the probability that we see at least $2$ breakdowns. That is to say, that we saw $2,3,4,5,6,7,\dots$ number of breakdowns. In other words, that we did not see $0,$ or $1$ breakdowns.
That would be $\Pr(X\geq 2) = \Pr(X=2)+\Pr(X=3)+\Pr(X=4)+\dots$, or more easily calculatable it would be $\Pr(X\geq 2) = 1 - \Pr(X<2) = 1 - \Pr(X=0)-\Pr(X=1)$. That is to say, some times it is easier to calculate the probability of the opposite event and subtract away from $1$.
Note: Here, we were given $\lambda$ in terms of average number of breakdowns per year and we were asked questions about the number of breakdowns we actually see in a year... we used the same timeframe for both the given information of $\lambda$ as well as for the questions we were asking. This is important to note. You might in future problems need to rephrase things to get them to match. If for instance, we were asked the probability that we get $5$ breakdowns in a decade (a span of ten years) for instance, then $2.5$ being the number of breakdowns in year wouldn't work for our lambda, but it would instead become $\lambda = 25$ as that would be the average number of breakdowns in a decade. We needed to make sure the timeframes we were referring to matched.