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This is the very last question on an assignment sheet and for some reason I can't wrap my head around the very last sentence without doubting my approach. Here's the question:

Tony’s home has a number of electric lights, which fail from time to time and have to be replaced. On average Tony has to replace 1.5 light bulbs every month.

(i) What is the appropriate distribution of the total number of light bulbs which need to be replaced each year? What are its mean and variance?

(ii) What is the probability that Tony will have to replace at least one light bulb in each of the next three months?

For (i) I've said that I think it's Poisson Distribution and both the mean and variance will be 18 for a full year.

For (ii) I've written down the following:

Let $ \lambda$ be $1.5$, $P(X \ge 1) = 1 - P(X = 0)$

$$ P(X = 0) = {e^{1.5} \cdot 1.5^0 \over 0!}$$

$$= 1 - 0.3347\ldots$$

I'm not even sure if that's on the right track, but it occured to me that I have to somehow account for the last part:

in each of the next three months.

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Yes, thanks - I've been up all night watching Harvard lectures to try get ahead with my coursework so my brain is a bit fried. – rtheunissen Mar 11 '13 at 19:24
By the way, the numerical calculation of $\Pr(X=0)$ is not correct. This probability is $e^{-1.5}\approx 0.22313$. It looks as if you multiplied by $(1.5)^1$, not by $(1.5)^0$, which is $1$. – André Nicolas Mar 11 '13 at 19:27
Yeah you're right, silly mistake. But right now I'm really confused about when to use $1 - ...$. Because I see $Pr(X = x)$ and $Pr(X > x)$ and $Pr(X < x)$. When should I use which? – rtheunissen Mar 11 '13 at 20:20
The poisson takes on only non-negative integer values. Use $\Pr(X\gt x)$ is the wording is something like "is greater than $x$." Usually the English wording will be quite precise. If you want the probability that $X$ is at least $2$, that means $2$ or more, so $\gt 1$. Easier to get at as $1-(\Pr(X=0)+\Pr(X=1))$. – André Nicolas Mar 11 '13 at 22:10
But what if it says something like X is at least 600? Is there a general solution? – rtheunissen Mar 12 '13 at 7:50
up vote 1 down vote accepted

We are invited to assume that the lifetimes of bulbs have exponential distribution, which is a somewhat unreasonable assumption. But never mind, we will hold our noses and think Poisson.

The number of failures in a month has Poisson distribution, mean $1.5$.

We have $3$ independent Poisson random variables, $X_1$, $X_2$, and $X_3$, which give us the number of failures in the first month, the second, the third.

We want the probability that $X_1\gt 0$ and $X_2\gt 0$ and $X_3\gt 0$.

Note that $\Pr(X_i\gt 0)=1-e^{-1.5}$.

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Does the 'and' imply that I should multiply them together as they are independent? – rtheunissen Mar 11 '13 at 19:16
Yes. That's why the "and" was in boldface, so that it would remind you of what you know. – André Nicolas Mar 11 '13 at 19:19
One more thing, you mentioned that $\Pr(X_i\gt 0)=1-e^{-1.5}$, but isn't that for P ~ Expo?, ie the CDF – rtheunissen Mar 11 '13 at 19:23
The probability for the Poisson is the same (and for good reason). For the Poisson, the formula you are familiar with says $e^{-1.5}\frac{(1.5)^0}{0!}$. But $0!=1$ and $(1.5)^0=1$. – André Nicolas Mar 11 '13 at 19:31

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