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Updated on: 11 Jun 2015, 09:32
Originally posted by Nevernevergiveup on 11 Jun 2015, 09:29.
Last edited by Bunuel on 11 Jun 2015, 09:32, edited 1 time in total.
Last edited by Bunuel on 11 Jun 2015, 09:32, edited 1 time in total.
Renamed the topic and edited the question.
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
40% (02:38) correct
60%
(03:02)
wrong
based on 167
sessions
History
Date
Time
Result
Not Attempted Yet
The bear alarm at Grizzly’s Peak ski resort sounds an average of once every thirty days, but the alarm is so sensitively calibrated that it sounds an average of ten false alarms for every undetected bear. Despite this, the alarm only sounds for three out of four bears that actually appear at the resort. On any given day at the resort, what is the approximate probability that there is neither an alarm nor an undetected bear?
(A) 224/365
(B) 13/14
(C) 14/15
(D) 376/390
(E) 29/30
(A) 224/365
(B) 13/14
(C) 14/15
(D) 376/390
(E) 29/30
14 Mar 2018, 15:39
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This is one of the more challenging questions you might see dealing with probability. Many people think that this question is confusing, but this confusion is often caused because people don't know how to think about the question. (In fact, think about it for a second: how often do we see GMAT questions that are deliberately worded in a way that confuses those who skim the problem too rapidly?) For example, there are those that believe we have to assume that only one "Bear" or "alarm" event happens per day. This is an unnecessary assumption. Since the problem talks about an "average" rate, we can think about all rates in these terms, without having to worry about the possibility of multiple events happening in a single day. Over time, multiple events in a single day would still be taken into account when we attempt to calculate an "average" rate.
Once you have a strategy that defines how you approach the question, it actually becomes very doable.
Here is the full "GMAT Jujitsu" for this question: 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, the solution to this problem naturally unfolds. (I call a such tables "Matrix Boxes" in my classes.)
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:
\[
\begin{matrix}
& \text{Bears} & \text{No Bears} & \text{Totals} \\
\text{Alarms} & \text{3} & \text{10} & \text{13} \\
\text{No Alarms} & \text{1} & \text{376} & \text{377} \\
\text{Totals} & \text{4} & \text{386} & \text{390}
\end{matrix}
\]
The problem wants to know what the “approximate” odds are on any given day that we have neither alarms nor undetected bears. The word “approximate” here is actually a TRAP designed to bait you down a wrong road. Just look down at the answer choices. These are very close together, and trying to approximate the answer would only cause us to miss the question. The problem asks, “what is the…probability that there is neither an alarm nor an undetected bear?” By using our Matrix Box, we can calculate this very quickly:
\(\frac{\text{No Alarms and No Bears}}{\text{Total Days}}=\frac{376}{390}\)
The answer is D.
By the way, for those of you that need extra practice, see these related questions:
https://gmatclub.com/forum/the-bear-ala ... l#p1386648
https://gmatclub.com/forum/the-bear-ala ... l#p2032644
https://gmatclub.com/forum/grizzly-peak ... l#p2033282
Once you have a strategy that defines how you approach the question, it actually becomes very doable.
Here is the full "GMAT Jujitsu" for this question: 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, the solution to this problem naturally unfolds. (I call a such tables "Matrix Boxes" in my classes.)
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:
\[
\begin{matrix}
& \text{Bears} & \text{No Bears} & \text{Totals} \\
\text{Alarms} & \text{3} & \text{10} & \text{13} \\
\text{No Alarms} & \text{1} & \text{376} & \text{377} \\
\text{Totals} & \text{4} & \text{386} & \text{390}
\end{matrix}
\]
The problem wants to know what the “approximate” odds are on any given day that we have neither alarms nor undetected bears. The word “approximate” here is actually a TRAP designed to bait you down a wrong road. Just look down at the answer choices. These are very close together, and trying to approximate the answer would only cause us to miss the question. The problem asks, “what is the…probability that there is neither an alarm nor an undetected bear?” By using our Matrix Box, we can calculate this very quickly:
\(\frac{\text{No Alarms and No Bears}}{\text{Total Days}}=\frac{376}{390}\)
The answer is D.
By the way, for those of you that need extra practice, see these related questions:
https://gmatclub.com/forum/the-bear-ala ... l#p1386648
https://gmatclub.com/forum/the-bear-ala ... l#p2032644
https://gmatclub.com/forum/grizzly-peak ... l#p2033282
General Discussion
11 Jun 2015, 12:38
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I find the wording of this question very confusing, for quite a few reasons. We need to make some assumptions unmentioned in the question to get the answer (for example, we need to assume that alarms never go off on the same day that a bear sneaks past the alarm system). And I had to read the part about how "sensitively calibrated" the alarm is a few times, because one wouldn't measure that by comparing false alarms to 'undetected bears'. Rather one would compare false alarms to detected bears. And I think it's problematic to ask for an "approximate probability", and then offer answer choices that differ by about 0.0025 (376/390 and 29/30 are only 0.0025 apart). Both answers give the "approximate probability" that the event happens, because both answers are nearly the same value.
So I don't think anyone should be concerned if they find the wording confusing. But to solve, you can imagine 4 bears show up. The alarm detects 3 of them, and does not detect 1 of them. We know for every 1 bear the alarm does not detect, the alarm goes off 10 times. So the alarm goes off 3 times when there is a bear, and 10 times when there is not. That is, every 13 times the alarm goes off, there is 1 bear that was not detected.
We know the alarm goes off once every 30 days, so the alarm would go off 13 times in 390 days. We just found out that there will be 1 undetected bear if the alarm goes off 13 times. So in a 390 day period, the alarm will go off 13 times, and there will be 1 undetected bear, so on 14 days there will be an alarm or an undetected bear, and thus on 376 days there will be neither. So the answer is 376/390.
So I don't think anyone should be concerned if they find the wording confusing. But to solve, you can imagine 4 bears show up. The alarm detects 3 of them, and does not detect 1 of them. We know for every 1 bear the alarm does not detect, the alarm goes off 10 times. So the alarm goes off 3 times when there is a bear, and 10 times when there is not. That is, every 13 times the alarm goes off, there is 1 bear that was not detected.
We know the alarm goes off once every 30 days, so the alarm would go off 13 times in 390 days. We just found out that there will be 1 undetected bear if the alarm goes off 13 times. So in a 390 day period, the alarm will go off 13 times, and there will be 1 undetected bear, so on 14 days there will be an alarm or an undetected bear, and thus on 376 days there will be neither. So the answer is 376/390.
