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Despite taking several statistics courses in college years ago, I've never had a mind for probabilities, so please excuse what may be a very basic question.

To make this fun, let's say you are in a haunted house and I am giving probabilities on encountering ghosts, room by room. Here are the probabilities:

  • 1/12 chance of encountering exactly 1 ghost in the dining room
  • 1/16 chance of encountering exactly 2 ghosts in the kitchen
  • 1/4 chance of encountering exactly 1 ghost in the bedroom
  • 1/30 chance of encountering either exactly 2 ghosts or exactly 8 ghosts in the cellar
  • 1/36 chance of encountering either exactly 3 ghosts or exactly 18 ghosts in the library

Probabilities are all or nothing. In the dining room, for example, there is an 11/12 probability of encountering zero ghosts.

These probabilities are independent of each other (e.g., encountering a ghost in one room does not affect whether or not you will encounter a ghost in another room).

Here are my questions (assume you enter every room):

  1. What is the probability you will encounter at least one ghost?
  2. Including upper and lower bounds, how many ghosts can you expect to encounter, if any?

Abstractions are helpful, but please also solve the problem with the concrete values above so I have an easier time mapping your work to an abstracted solution. Thank you!

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Is the chance of encountering exactly $1$ ghost in the kitchen equal to $0$? Is the chance of encountering $2$ ghosts in the cellar the same as that of encountering $3$? Is that chance in both cases $1 \over 30$ or $1 \over {7 \cdot 30}$ (so that the chance of encountering any ghosts sums up to $1 \over 30$)? – Karolis Juodelė Mar 17 '13 at 19:56
In the kitchen you have a 1/16 chance of encountering exactly 2 ghosts and a 15/16 chance of encountering 0 ghosts. As for the ranges of ghosts, I'm really only looking for a calculation that includes 2 ghosts in the cellar and 3 ghosts in the library and another that includes 8 ghosts in the cellar and 18 ghosts in the library (the upper and lower bounds I was referring to). – Matthew Ratzloff Mar 17 '13 at 20:00
I've updated the question to be more exacting with my language. – Matthew Ratzloff Mar 17 '13 at 20:09
up vote 2 down vote accepted

$$\begin{align} P(\text{at least one ghost}) &= 1 - P(\text{no ghosts}) \\&= 1 - P(\text{no ghost in dining room}) \cdot P(\text{no ghost in kitchen}) \cdots \\&= 1 - \Big (1-\frac{1}{12} \Big ) \Big (1-\frac{1}{16} \Big ) \Big(1-\frac{1}{4}\Big)\cdots\end{align}$$

$$\begin{align} N_\text{expected} &= N_\text{dining room} \cdot P(\text{dining room}) + N_\text{kitchen} \cdot P(\text{kitchen}) + \dots \\&= 1\cdot \frac{1}{12} + 2 \cdot \frac{1}{16} + 1 \cdot \frac{1}{4} + \dots \end{align}$$

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Excellent. I had the right answer to N{expected}, but P(at least one ghost) was eluding me. Thanks for your help! – Matthew Ratzloff Mar 17 '13 at 20:55

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