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The number of cars $(X)$ arriving at a service station per day follows a Poisson distribution with mean $4$. The service station can provide service to a maximum of $4$ cars per day. Then the expected number of cars that do not get service per day equals

1. $4$
2. $0$
3. $\Sigma_{i=0}^{\infty} i P(X=i+4)$
4. $\Sigma_{i=4}^{\infty} i P(X=i-4)$

### 1 comment

C?

$\mathbf{\underline{Answer:\Rightarrow C}}$

$\mathbf{\underline{Explanation:\Rightarrow}}$

If $\mathbf{X = 0}$, then no cars arrive and the number of cars that do not get service $\mathbf{ = 0}$

If $\mathbf{X = 1}$, then no cars arrive and the number of cars that do not get service $\mathbf{ = 0}$

If $\mathbf{X = 2}$, then no cars arrive and the number of cars that do not get service $\mathbf{ = 0}$

If $\mathbf{X = 3}$, then no cars arrive and the number of cars that do not get service $\mathbf{ = 0}$

If $\mathbf{X = 4}$, then no cars arrive and the number of cars that do not get service $\mathbf{ = 0}$

If $\mathbf{X = 5}$, then no cars arrive and the number of cars that do not get service $\mathbf{ = 1}$

If $\mathbf{X = 6}$, then no cars arrive and the number of cars that do not get service $\mathbf{ = 2}$

$\mathbf{\vdots} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \vdots\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \mathbf{\vdots}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \vdots\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \vdots$

$\therefore \text{The expected number of cars missing out on service} \\\\= \sum(\text{number of cars that miss out being serviced})(\text{probability of this happening}) \\\\=\mathbf{ 0P(X = 0) + 0P(X = 1) + 0P(X = 2)+\cdots+1P(X = 5) + 2P(X = 6)+\cdots\cdots\cdots}$

So, the above expression can be represented in the compact form as:

$$\sum(\text{The number of cars that miss-out being serviced})(\text{Probabibility of this happening}) = \mathbf{\sum_{0}^{\infty}iP(X = i+4)}$$

$\therefore$ Option $\mathbf C$ is the right answer.
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