I have a probability distribution as follows, where X denotes the number of accidents $$ P(X=i) = K \cdot \frac{2^i}{i!}, \quad \quad i=0,1,2,...$$ and I need to find out what the value of the positive constant K is. Usually to calculate a constant in the PDF, I integrate the function and equal it to $1$, so I can calculate the constant. But in this case, how can I integrate the function? I don't think it's possible since there's the $i!$, so maybe it has something to do with the $E(X)$ instead? So I guess it is something like $$E(X) = \sum_{i=0}^\infty i\cdot K \cdot \frac{2^i}{i!}$$ I'm not sure where to go on from here though. Also, the second part asks for the probability of having 3 or more accidents, which I believe means calculate $P(X \ge 3)$. How do I calculate this? I don't think it should have anything to do with gamma functions.$\qquad \qquad \qquad \qquad \qquad \qquad \qquad$ I'd really appreciate your help. Thanks in advance!


All you need to do is make sure that $\sum_n P(X = n) = 1$, that is

$$ 1 = \sum_n k \frac{2^n}{n!} = k\sum_n \frac{2^n}{n!} = ke^2 $$

so $k = e^{-2}$. In this last step I used

$$ e^x = \sum_n \frac{x^n}{n!} $$

As for the second question

\begin{eqnarray} P(X\ge 3) &=& \sum_{n\ge 3}P(X = n) = \left(-\sum_{n < 3}P(X = n) + \sum_{n < 3}P(X = n)\right) + \sum_{n\ge 3}P(X = n) \\ &=& -\sum_{n < 3}P(X = n) + \sum_{n \ge 0}P(X = n) = 1 -\sum_{n < 3}P(X = n) \\ &=& 1 - P(X = 0) - P(X= 1) - P(X=2) = 1 - e^{-2}\left(1 + 2 + \frac{2^2}{2!} \right) \\ &=& = 1 - 5e^{-2}\approx 0.32 \end{eqnarray}

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