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In a particular city, the probability a call to a fire department concerns various situations is as given below:

  1. fire in a detached home: $p_1=0.10$
  2. fire in a semi-detached home: $p_2=0.05$
  3. fire in an apartment or multiple unit residence: $p_3=0.05$
  4. fire in a non-residential building: $p_4=0.15$
  5. non-fire-related emergency: $p_5=0.15$
  6. false alarm: $p_6=0.50$

(a) Give the joint probability function for $X_1,\dots,X_6$ (I know how to do this one...)

(b) What is the probability that there is at least one apartment fire, given that there are 4 fire-related calls? (cannot come up with a formula for this one)

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Try this: in 100 calls, how many are fire-related, and how many of those are apartment fire calls? So what is the probability of non-apartment fire call? Of four of them? And what's the complement of "none"? –  User58220 Mar 24 '13 at 1:26
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1 Answer

up vote 2 down vote accepted

On average, in 100 calls there will be 50 false alarms and 15 non-fire-related calls.

So there will be only 35 fire related calls, of which 5 will be apartment calls. So the probability that a fire call will be an apartment call is $\frac{1}{7}$

The probability that a fire call will not be apartment related is $\frac{6}{7}$

The probability of four non-apartment fire calls is $(\frac{6}{7})^4$

So the probability of at least one apartment call in four fire calls is:$$1-(\frac{6}{7})^4$$ or $ 0.460225$

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