A car factory carries out a final inspection for scratches incurred during the manufacturing process. The inspectors have found that these scratches appear randomly around the cars (according to a Poisson process) with an average of 1.2 scratches per car.
Four of these cars have been selected at random to ship off to a new dealer. What is the probability that there are no more than 6 scratches in total on these four cars?
So I'm having some trouble understanding what is what in this question.
So far I have figured that I am trying to find the probability that event $X$ (car chosen has a scratch) is $\le6$ when $k$ (the number of cars chosen) $=4$.
However I am unsure which distribution formula to use. I have a feeling it is negative binomial however my formual only allows for $X$ actually equalling a number x, rather than being greater than or less than x (x being 6 in this case). The negative binomial formula I am using is below.
$$Pr(X=x) = \left(\frac{x-1}{k-1}\right)p^k(1-p)^{x-k}$$
So my questions are:
Am I using the right distribution for the question?
What letter/symbol does the scratches per car value take in the formula (the 1.2 value)?
Is my assumption correct in regards to my designation of the other letters/symbols (X = car chosen has a scratch, k = the number of cars chosen, x = 6)?
If the number of scratches on one car is Poisson distributed with a mean of $\lambda = 1.2$ scratches, then the total number of scratches on four cars, assuming that the number of scratches on each car is independent of the others and are all identically distributed, is simply $$S = X_1 + X_2 + X_3 + X_4,$$ where $X_i \sim \operatorname{Poisson}(\lambda).$ What is the sum of four such Poisson random variables? It is not difficult to show in the general case if $$A \sim \operatorname{Poisson}(\lambda_1), \quad B \sim \operatorname{Poisson}(\lambda_2)$$ are independent, $$\begin{align*} \Pr[A+B = k] &= \sum_{m=0}^k \Pr[A = m] \Pr[B = k-m] \\ &= \sum_{m=0}^k e^{-\lambda_1} \frac{\lambda_1^m}{m!} e^{-\lambda_2} \frac{\lambda_2^{k-m}}{(k-m)!} \\ &= e^{-(\lambda_1 + \lambda_2)} \frac{1}{k!} \sum_{k=0}^m \binom{k}{m} \lambda_1^k \lambda_2^{k-m} \\ &= e^{-(\lambda_1 + \lambda_2)} \frac{(\lambda_1 + \lambda_2)^k}{k!}, \end{align*}$$ where the last step follows from the binomial theorem. Thus $A+B \sim \operatorname{Poisson}(\lambda_1 + \lambda_2)$, and it is obvious from induction that in the above case, we must have $$S \sim \operatorname{Poisson}(4\lambda).$$ Therefore, the distribution of the total number of scratches on the four cars is Poisson with rate $4\lambda = 4.8$, and the probability that there are no more than 6 scratches is simply $$\Pr[S \le 6] = \sum_{k=0}^6 e^{-4\lambda} \frac{(4\lambda)^k}{k!}.$$
An alternative proof of the distribution of $S$ can be obtained through the use of moment generating functions. The Poisson MGF is $$\begin{align*} M_X(t) &= \operatorname{E}[e^{tX}] \\ &= \sum_{k=0}^\infty e^{tk} \Pr[X = k] \\ &= e^{-\lambda} \sum_{k=0}^\infty \frac{(e^t \lambda)^k}{k!} \\ &= \exp(-\lambda + e^t \lambda) \sum_{k=0}^\infty \exp(-e^t \lambda) \frac{(e^t \lambda)^k}{k!} \\ &= \exp(-\lambda(1-e^t)), \end{align*}$$ the last sum being the sum over a Poisson distribution with rate parameter $\lambda^* = e^t \lambda$, which equals $1$. Now a property of MGFs is that $$M_S(t) = (M_X(t))^n,$$ if $S = X_1 + X_2 + \cdots + X_n$ and $X_i \sim X$ are iid. Then we obtain $$M_S(t) = \exp(-n \lambda(1-e^t))$$ which immediately corresponds to a Poisson MGF with parameter $n\lambda$; hence $S \sim \operatorname{Poisson}(n\lambda)$ if $X_i \sim \operatorname{Poisson}(\lambda)$, by uniqueness of MGFs.