Russell's Probability Question Please take a look at the following question
The bear alarm at resort sounds an average of once every 30 days, but the alarm is so sensitively calibrated that is sounds an average of 10 false alarm for every undetected bear. Despite this, the alarm only sounds for 3 out of 4 bears that actually appear at the resort. 
 1- if the alarm sounds, what is the probability that a bear has actually been sighted? 
 2_ on any given day at resort, what is the approximate probability there is neither an alarm nor an undetected bear?
 3 _ if the alarm were to sound an average of 10 false alarms for every detected bear, the probability that a sounded alarm would indicate an actual bear would
 1- decrease by 75% 2_ increase by 33% 3_ increase by 75% 4- decrease by 33% 5_ decrease by 14%
4- approximately how many bears at the resort each year?
 A: Consider an "incident" to occur when either a bear appears or an alarm sounds, or both.
I'll suppose incidents occur as a Poisson process, with rate $r$ per day.  Each incident may be of any of the following $3$ types: 


*

*A bear appears and there is an alarm

*A bear appears but there is no alarm

*A false alarm with no bear


I'll suppose, independent of all other incidents, each incident has probabilities $p_1, p_2, p_3$ respectively of being each of these types.  Thus $p_1 + p_2 + p_3 = 1$.  We are told that $p_3 = 10 p_2$ (10 false alarms for each undetected bear) and $p_1/(p_1 + p_2) = 3/4$ (alarm sounds for 3 out of 4 bears that appear).  From this we get $p_1 = 3/14$, $p_2 = 1/14$, $p_3 = 10/14$.  This should help you answer the first question.
Now the rate at which the alarm sounds (type 1 or 3) is $(p_1 + p_3) r = 13 r/14$.
The "alarm sounds once out of 30 days" says this is $1/30$ per day, so $r = 14/390$.  This should help you answer the second question.
A: Answer
Part I 
P(Alarm sounds for undetected Bear) =   10/11
P(Alarm does not sound for undetected Bear) = 1/11
P(Alarm sounds for detected Bear) = 3/4
P(Alarm does not sound for detected Bear) = 1/4
P(Detected Bear/Alarm sounds) = $\dfrac{\frac{3}{4}}{\frac{3}{4}+\frac{10}{11}} = \frac{33}{73}$
P(No alarm and undetected Bear) =$\dfrac{\frac{1}{11}}{\frac{1}{4}+\frac{1}{11}} = \frac{4}{15}$
Rate would be $\frac{4}{15}*\frac{1}{30} = \frac{4}{450}$
Part III
P(Alarm sounds for undetected Bear) =   10/11
P(Alarm does not sound for undetected Bear) = 1/11
P(Alarm sounds for detected Bear) = 1/11
P(Alarm does not sound for detected Bear) = 10/11
P(Detected Bear/Alarm sounds) = $\dfrac{\frac{1}{11}}{\frac{1}{11}+\frac{10}{11}} = \frac{1}{11}$
Decrease by ($\frac{33}{73}$ - $\frac{1}{11})$/($\frac{33}{73})$ = 79%
PART IV
Total Number of Bears from Part I
=365/30 = $\frac{365}{30}$
=12.16 Bears.
