The bear alarm at Grizzly’s Peak ski resort sounds an averag

Thank you for your request! I would be very glad to help. There have been some great abbreviated responses by others to this question, but several people (including a couple of my own students) have asked for a more complete solution, so I thought it would be useful to post a solution on this forum. For those of you preparing for the GMAT, it is important to note here that while it is useful to learn how to solve a specific question, it is more useful to learn how to THINK about whole classes of questions. That is how I am going to approach my explanation.

There are actually two totally different ways to solve this problem: (1) Using a combination of rates and ratios to arrive at the answer, (2) Using a specialized table I call in my classes a "Matrix Box". Here is the full "GMAT Jujitsu" for this question, using both techniques to arrive at the same answer. For those of you preparing for the GMAT, pick which ever technique resonates with you the most.

Technique #1: Using Rates and Ratios

The initial probability is fairly easy to calculate. First, the problem tells us that there are ten false alarms for every undetected bear. Thus,

\(\frac{\text{False Alarms}}{\text{Undetected Bears}}=\frac{10}{1}\)

The problem also tells us that the alarm sounds for three out of four bears that appear. (In other words, 3 bears are detected for every 1 undetected bear.) Thus,

\(\frac{\text{Detected Bears}}{\text{Undetected Bears}}=\frac{3}{1}\)

With these two ratios, it is very easy to see that the ratio between False Alarms and Detected Bears is 10:3, and that the total amount of alarms would be 10x + 3x = 13x.

Thus, the odds of a detected bear if the alarm sounds would be:

\(\frac{\text{Detected Bears}}{\text{Total Alarms}}=\frac{3x}{13x}=\frac{3}{13}\)

The answer is B.

Technique #2: Using Matrix Boxes

We have two mutually exclusive groups (detected vs. undetected bears and alarms vs. no alarms.) If you can organize your thoughts around these mutually-exclusive groups by creating a specialized table called a Matrix Box, the solution to this problem naturally unfolds.

Here is what the blank table might look like:
\[
\begin{matrix}
& \text{Bears} & \text{No Bears} & \text{Totals} \\
\text{Alarms} & \text{____} & \text{____} & \text{____} \\
\text{No Alarms} & \text{____} & \text{____} & \text{____} \\
\text{Totals} & \text{____} & \text{____} & \text{____}
\end{matrix}
\]
Matrix boxes are simplest if we can use concrete numbers. The problem (or the opportunity) here is that we are only given ratios. This means we can invent our own total amount, using leverage inside the problem to guess what number would work the best. The problem states that an alarm sounds "an average of once every thirty days", so our total should be at least a multiple of 30 to avoid messy fractions. But we also want to have it be a multiple of the “total alarms” in order to make the math easy. It looks like the ratios involving total alarms have false alarms in ratios of 10 and real alarms in ratios of 3. So, total alarms should be at least in ratios of 13. Thus, we want a number that is a multiple of \(13\) and a multiple of \(30\). Thus, if we were to pick \(13*30=390\) as our total value, we wouldn't have to deal with ugly math. (Notice, too that \(390\) is the denominator in answer choice D, which might give us a hint here.)

Let's start there:
\[
\begin{matrix}
& \text{Bears} & \text{No Bears} & \text{Totals} \\
\text{Alarms} & \text{____} & \text{____} & \text{____} \\
\text{No Alarms} & \text{____} & \text{____} & \text{____} \\
\text{Totals} & \text{____} & \text{____} & \text{390}
\end{matrix}
\]
The problem tells us that we see 1 alarm happening about every 30 days. Thus over 390 days we get 13 alarms. (We can plug this in to the “Total alarms” slot.) The next two ratios tell us that we get 10 false alarms for every undetected bear (or failed alarm) and that 3 bears are detected for every one bear that is not detected (again, a failed alarm). Notice the “1 undetected bear” is common to both. Plugging all of these values into our Matrix box gives us:
\[
\begin{matrix}
& \text{Bears} & \text{No Bears} & \text{Totals} \\
\text{Alarms} & \text{3} & \text{10} & \text{13} \\
\text{No Alarms} & \text{1} & \text{____} & \text{____} \\
\text{Totals} & \text{____} & \text{____} & \text{390}
\end{matrix}
\]
It is now very easy to fill in the rest of the box:
\[
\begin{matrix}
& \text{Bears} & \text{No Bears} & \text{Totals} \\
\text{Alarms} & \text{3} & \text{10} & \text{13} \\
\text{No Alarms} & \text{1} & \text{376} & \text{377} \\
\text{Totals} & \text{4} & \text{386} & \text{390}
\end{matrix}
\]
Of course, the target of this question doesn't even require us to fill in the whole table. (I am including the table here to show how quickly you can leverage a Matrix Box to solve for any number of possible variables or ratios.) In the end,we only need a few numbers from this table. (And this information we can extract from the table without actually filling in all the blanks on the table!) The problem just wants to know “If the alarm sounds, what is the probability that a bear has actually been sighted?” By using our Matrix Box, we can calculate this very quickly:

\(\frac{\text{Alarms with Bears}}{\text{Total Alarms}}=\frac{3}{13}\)

No matter how you solve it, the answer is still B.

For those of you that need extra practice, see these related questions:
https://gmatclub.com/forum/the-bear-ala ... l#p2032644
https://gmatclub.com/forum/grizzly-peak ... l#p2033282
https://gmatclub.com/forum/the-bear-ala ... l#p1536872