For a Poisson distribution with mean $\lambda$ and a capacity of $c$:
The number of cars not served is $\max(X-c, 0)$.
The expected number of cars not served is $E[\max(X-c, 0)]$.
$$E[\max(X-c, 0)] = \sum_{x=0}^{\infty} \max(x-c, 0) \cdot P(X=x)$$
Since $\max(x-c, 0)$ is non-zero only for $x>c$, the sum becomes:
$$\sum_{x=c+1}^{\infty} (x-c) \cdot P(X=x)$$
Let $i=x-c$. As $x$ goes from $c+1$ to $\infty$, $i$ goes from $1$ to $\infty$. Also, $x=i+c$.
So, the expression becomes $\sum_{i=1}^{\infty} i \cdot P(X=i+c)$.
Since the term for $i=0$ is $0 \cdot P(X=0+c)=0$, we can start the sum from $i=0$.
The general formula is $\sum_{i=0}^{\infty} i \cdot P(X=i+c)$.
In this problem, $\lambda=4$ and $c=4$.
Substituting $c=4$ into the formula, we get:
$$\sum_{i=0}^{\infty} i \cdot P(X=i+4)$$
This matches option C.
1. Identify the "Excess":
The number of items not processed is the number of arrivals above the capacity.
This is mathematically represented as $\max(X-c, 0)$.
If $X \leq c$, the excess is $0$.
If $X > c$, the excess is $X-c$.
2. Formula for Expected Excess:
The expected value, $E[\text{Excess}]$, is the sum of (excess amount) $\times$ (its probability).
$E[\text{Excess}] = \sum_{x=0}^{\infty} \max(x-c, 0) \cdot P(X=x)$.
Since the excess is only non-zero when $x>c$, the sum simplifies to:
$$E[\text{Excess}] = \sum_{x=c+1}^{\infty} (x-c) \cdot P(X=x)$$
3. Pattern Matching:
You don't need to derive the formula every time.
The general form for the expected excess with capacity $c$ is:
$$\sum_{i=0}^{\infty} i \cdot P(X=i+c)$$
Memorize this pattern. The terms in the sum represent:
$i=1: 1 \cdot P(X=c+1)$ (1 excess car)
$i=2: 2 \cdot P(X=c+2)$ (2 excess cars)
and so on...
Just remember that the expected excess for a capacity $c$ is a summation of terms like $i \cdot P(X=i+c)$, starting from $i=0$ or $i=1$. In this case, with $c=4$, you look for $\sum i \cdot P(X=i+4)$. This saves time and avoids calculation errors.
