0
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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}

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.