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19 Mar 2018, 16:03
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C
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:
27% (02:12) correct
73%
(02:29)
wrong
based on 142
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. If the alarm were to sound an average of ten false alarms for every detected bear, the probability that a sounded alarm would indicate an actual bear would
A) Decrease by approximately 75%
B) Decrease by approximately 33%
C) Decrease by approximately 14%
D) Increase by approximately 33%
E) Increase by approximately 75%
A) Decrease by approximately 75%
B) Decrease by approximately 33%
C) Decrease by approximately 14%
D) Increase by approximately 33%
E) Increase by approximately 75%
Here is the full "GMAT Jujitsu" for this question:
First, this problem is a little tricky, because it sounds like a percentage change question. However, if you read it carefully, it focuses on a change in a percentage, not an official percentage change. The question asks how the probability "that a sounded alarm would indicate an actual bear" would change "if the alarm were to sound an average of ten false alarms for every detected bear." 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, our initial odds are:
\(P_{initial}=\frac{\text{Detected Bears}}{\text{Total Alarms}}=\frac{3x}{13x}=\frac{3}{13}\)
Now we need to calculate the hypothetical change in alarm rates. From the question, we know that \(\frac{\text{False Alarms}}{\text{Detected Bears}}=\frac{10}{1}\)
Using the same logic as above, we can leverage this relationship to determine the ratio of Detected Bears to Total Alarms:
\(P_{new}=\frac{\text{Detected Bears}}{\text{Total Alarms}}=\frac{\text{Detected Bears}}{\text{False Alarms + Detected Bears}}=\frac{1}{(10+1)}=\frac{1}{11}\)
The difference between these two probabilities would be:
\(P_{initial}-P_{new}=\frac{3}{13}-\frac{1}{11}\)
Now, don't fall for the trap and actually try to do this math. Remember: you don't have a calculator on the Quant portion of the GMAT. (In fact, those who claim simply that \(3/13-1/11 = 0.1398\) are either using a calculator or doing the math the long way around!) The GMAT often baits inexperienced test-takers into trying to solve very messy math, when looking down at the answer choices would be sufficient to answer the question. We can leverage the answer choices against each other and look for the approximate value. It is easy to see that \(\frac{3}{13} > \frac{1}{11}\), so the probability is decreasing (thereby eliminating answer choices D and E.) But a probability can’t decrease below \(0\), and \(\frac{3}{13}\) is clearly less than \(75\%\) or \(33\%\). (Think about this: \(\frac{3}{9}\) is \(33\%\), so \(\frac{3}{13}\) is considerably smaller than that.) The only possible value it could be is \(14\%\).
You can also just do the math and approximate. \(\frac{3}{13} ≈ 23\%\) and \(\frac{1}{11}≈ 9\%\). This means the probability dropped about \(14\%\).
No matter how you look at it, the answer is C.
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/grizzly-peak ... l#p2033282
https://gmatclub.com/forum/the-bear-ala ... l#p1536872
First, this problem is a little tricky, because it sounds like a percentage change question. However, if you read it carefully, it focuses on a change in a percentage, not an official percentage change. The question asks how the probability "that a sounded alarm would indicate an actual bear" would change "if the alarm were to sound an average of ten false alarms for every detected bear." 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, our initial odds are:
\(P_{initial}=\frac{\text{Detected Bears}}{\text{Total Alarms}}=\frac{3x}{13x}=\frac{3}{13}\)
Now we need to calculate the hypothetical change in alarm rates. From the question, we know that \(\frac{\text{False Alarms}}{\text{Detected Bears}}=\frac{10}{1}\)
Using the same logic as above, we can leverage this relationship to determine the ratio of Detected Bears to Total Alarms:
\(P_{new}=\frac{\text{Detected Bears}}{\text{Total Alarms}}=\frac{\text{Detected Bears}}{\text{False Alarms + Detected Bears}}=\frac{1}{(10+1)}=\frac{1}{11}\)
The difference between these two probabilities would be:
\(P_{initial}-P_{new}=\frac{3}{13}-\frac{1}{11}\)
Now, don't fall for the trap and actually try to do this math. Remember: you don't have a calculator on the Quant portion of the GMAT. (In fact, those who claim simply that \(3/13-1/11 = 0.1398\) are either using a calculator or doing the math the long way around!) The GMAT often baits inexperienced test-takers into trying to solve very messy math, when looking down at the answer choices would be sufficient to answer the question. We can leverage the answer choices against each other and look for the approximate value. It is easy to see that \(\frac{3}{13} > \frac{1}{11}\), so the probability is decreasing (thereby eliminating answer choices D and E.) But a probability can’t decrease below \(0\), and \(\frac{3}{13}\) is clearly less than \(75\%\) or \(33\%\). (Think about this: \(\frac{3}{9}\) is \(33\%\), so \(\frac{3}{13}\) is considerably smaller than that.) The only possible value it could be is \(14\%\).
You can also just do the math and approximate. \(\frac{3}{13} ≈ 23\%\) and \(\frac{1}{11}≈ 9\%\). This means the probability dropped about \(14\%\).
No matter how you look at it, the answer is C.
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/grizzly-peak ... l#p2033282
https://gmatclub.com/forum/the-bear-ala ... l#p1536872
General Discussion
20 Mar 2018, 12:57
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Bookmarks
for the 3 detected bears there were 3 alarms and for the 1 undetected bear there was 10 alarms making the total # of alarms to be 13 per 3 detected bears.
If the ratio between false alarms and detected bears were 10:1 that means that there are 11 alarms per 1 detected bear.
3/13-1/11 = 0.1398 ~ 14%
If the ratio between false alarms and detected bears were 10:1 that means that there are 11 alarms per 1 detected bear.
3/13-1/11 = 0.1398 ~ 14%
