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27 May 2024, 01:30
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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:
75%
(hard)
Question Stats:
50% (01:46) correct
50%
(01:57)
wrong
based on 133
sessions
History
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Not Attempted Yet
An airport screens bags for forbidden items and an alarm is supposed to be triggered when a forbidden item is detected. Suppose that 5% of bags contain forbidden items. Given a randomly chosen bag triggers the alarm, what is the probability that it contains a forbidden item?
(1) If a bag contains a forbidden item, there is a 98% chance that it triggers the alarm.
(2) If a bag doesn't contain a forbidden item, there is an 8% chance that it triggers the alarm.
(1) If a bag contains a forbidden item, there is a 98% chance that it triggers the alarm.
(2) If a bag doesn't contain a forbidden item, there is an 8% chance that it triggers the alarm.
27 May 2024, 03:38
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We are told that 5% of bags contain a forbidden item, therefore 95% do not.
Given the phrasing of the question, one can conclude that there are four possible outcomes which can occur:
1) A banned item triggers the alarm (Let that be represented by BT)
2) A banned item that does not trigger the alarm (Let that be represented by BnT)
3) A non-banned item which triggers the alarm (Let that be represented by nBT)
4) A non-banned item which does not trigger the alarm (Let that be represented by nBnT)
The probability that a triggered alarm is a banned item will be (pBT)/(pBT + pnBT) [where p represents probability]
(1) If a bag contains a forbidden item, there is a 98% chance that it triggers the alarm.
From this statement one can work out the pBT. However, without pnBt one cannot solve this question.
INSUFFICIENT
(2) If a bag doesn't contain a forbidden item, there is an 8% chance that it triggers the alarm.
From this statement one can work out the pnBT. However, without pBt one cannot find the probability that a randomly chosen bag that triggers the alarm will contain a banned item.
INSUFFICIENT
(1+2)
Putting the statements together will yield values for both pBT and for pnBT and thus one will be able to fill in the equation (pBT)/(pBT + pnBT) and have the probability that a randomly chosen bag that triggers the alarm will contain a banned item.
SUFFICIENT
ANSWER C
THE MATH:
(1) If a bag contains a forbidden item, there is a 98% chance that it triggers the alarm.
\(\frac{5}{100}*\frac{98}{100}\)
\(\frac{1}{20}*\frac{49}{50} = \frac{49}{1000}\) which is BT.
(2) If a bag doesn't contain a forbidden item, there is an 8% chance that it triggers the alarm.
\(\frac{8}{100}*\frac{95}{100}\)
\(\frac{2}{25}*\frac{19}{20} = \frac{19}{250}*\frac{4}{4} = \frac{76}{1000}\) IS nBT
(1+2)
Using the above values for BT and nBT in the equation: (pBT)/(pBT + pnBT)
\(\frac{\frac{49}{1000}}{\frac{49}{1000}+\frac{76}{1000}}\)
\(\frac{\frac{49}{1000}}{\frac{125}{1000}}\)
\(\frac{49}{1000}*\frac{1000}{125}\)
\(\frac{49}{125}\) [multiply through by \(\frac{8}{8}\)]
\(\frac{392}{1000}\)
Given the phrasing of the question, one can conclude that there are four possible outcomes which can occur:
1) A banned item triggers the alarm (Let that be represented by BT)
2) A banned item that does not trigger the alarm (Let that be represented by BnT)
3) A non-banned item which triggers the alarm (Let that be represented by nBT)
4) A non-banned item which does not trigger the alarm (Let that be represented by nBnT)
The probability that a triggered alarm is a banned item will be (pBT)/(pBT + pnBT) [where p represents probability]
(1) If a bag contains a forbidden item, there is a 98% chance that it triggers the alarm.
From this statement one can work out the pBT. However, without pnBt one cannot solve this question.
INSUFFICIENT
(2) If a bag doesn't contain a forbidden item, there is an 8% chance that it triggers the alarm.
From this statement one can work out the pnBT. However, without pBt one cannot find the probability that a randomly chosen bag that triggers the alarm will contain a banned item.
INSUFFICIENT
(1+2)
Putting the statements together will yield values for both pBT and for pnBT and thus one will be able to fill in the equation (pBT)/(pBT + pnBT) and have the probability that a randomly chosen bag that triggers the alarm will contain a banned item.
SUFFICIENT
ANSWER C
THE MATH:
(1) If a bag contains a forbidden item, there is a 98% chance that it triggers the alarm.
\(\frac{5}{100}*\frac{98}{100}\)
\(\frac{1}{20}*\frac{49}{50} = \frac{49}{1000}\) which is BT.
(2) If a bag doesn't contain a forbidden item, there is an 8% chance that it triggers the alarm.
\(\frac{8}{100}*\frac{95}{100}\)
\(\frac{2}{25}*\frac{19}{20} = \frac{19}{250}*\frac{4}{4} = \frac{76}{1000}\) IS nBT
(1+2)
Using the above values for BT and nBT in the equation: (pBT)/(pBT + pnBT)
\(\frac{\frac{49}{1000}}{\frac{49}{1000}+\frac{76}{1000}}\)
\(\frac{\frac{49}{1000}}{\frac{125}{1000}}\)
\(\frac{49}{1000}*\frac{1000}{125}\)
\(\frac{49}{125}\) [multiply through by \(\frac{8}{8}\)]
\(\frac{392}{1000}\)
09 Feb 2025, 08:30
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Bookmarks
Hi
I didn't understand the highlighted part. Cud u pls elaborate?
Thanks
I didn't understand the highlighted part. Cud u pls elaborate?
Thanks
Nidzo
We are told that 5% of bags contain a forbidden item, therefore 95% do not.
Given the phrasing of the question, one can conclude that there are four possible outcomes which can occur:
1) A banned item triggers the alarm (Let that be represented by BT)
2) A banned item that does not trigger the alarm (Let that be represented by BnT)
3) A non-banned item which triggers the alarm (Let that be represented by nBT)
4) A non-banned item which does not trigger the alarm (Let that be represented by nBnT)
The probability that a triggered alarm is a banned item will be (pBT)/(pBT + pnBT) [where p represents probability]
(1) If a bag contains a forbidden item, there is a 98% chance that it triggers the alarm.
From this statement one can work out the pBT. However, without pnBt one cannot solve this question.
INSUFFICIENT
(2) If a bag doesn't contain a forbidden item, there is an 8% chance that it triggers the alarm.
From this statement one can work out the pnBT. However, without pBt one cannot find the probability that a randomly chosen bag that triggers the alarm will contain a banned item.
INSUFFICIENT
(1+2)
Putting the statements together will yield values for both pBT and for pnBT and thus one will be able to fill in the equation (pBT)/(pBT + pnBT) and have the probability that a randomly chosen bag that triggers the alarm will contain a banned item.
SUFFICIENT
ANSWER C
THE MATH:
(1) If a bag contains a forbidden item, there is a 98% chance that it triggers the alarm.
\(\frac{5}{100}*\frac{98}{100}\)
\(\frac{1}{20}*\frac{49}{50} = \frac{49}{1000}\) which is BT.
(2) If a bag doesn't contain a forbidden item, there is an 8% chance that it triggers the alarm.
\(\frac{8}{100}*\frac{95}{100}\)
\(\frac{2}{25}*\frac{19}{20} = \frac{19}{250}*\frac{4}{4} = \frac{76}{1000}\) IS nBT
(1+2)
Using the above values for BT and nBT in the equation: (pBT)/(pBT + pnBT)
\(\frac{\frac{49}{1000}}{\frac{49}{1000}+\frac{76}{1000}}\)
\(\frac{\frac{49}{1000}}{\frac{125}{1000}}\)
\(\frac{49}{1000}*\frac{1000}{125}\)
\(\frac{49}{125}\) [multiply through by \(\frac{8}{8}\)]
\(\frac{392}{1000}\)
Given the phrasing of the question, one can conclude that there are four possible outcomes which can occur:
1) A banned item triggers the alarm (Let that be represented by BT)
2) A banned item that does not trigger the alarm (Let that be represented by BnT)
3) A non-banned item which triggers the alarm (Let that be represented by nBT)
4) A non-banned item which does not trigger the alarm (Let that be represented by nBnT)
The probability that a triggered alarm is a banned item will be (pBT)/(pBT + pnBT) [where p represents probability]
(1) If a bag contains a forbidden item, there is a 98% chance that it triggers the alarm.
From this statement one can work out the pBT. However, without pnBt one cannot solve this question.
INSUFFICIENT
(2) If a bag doesn't contain a forbidden item, there is an 8% chance that it triggers the alarm.
From this statement one can work out the pnBT. However, without pBt one cannot find the probability that a randomly chosen bag that triggers the alarm will contain a banned item.
INSUFFICIENT
(1+2)
Putting the statements together will yield values for both pBT and for pnBT and thus one will be able to fill in the equation (pBT)/(pBT + pnBT) and have the probability that a randomly chosen bag that triggers the alarm will contain a banned item.
SUFFICIENT
ANSWER C
THE MATH:
(1) If a bag contains a forbidden item, there is a 98% chance that it triggers the alarm.
\(\frac{5}{100}*\frac{98}{100}\)
\(\frac{1}{20}*\frac{49}{50} = \frac{49}{1000}\) which is BT.
(2) If a bag doesn't contain a forbidden item, there is an 8% chance that it triggers the alarm.
\(\frac{8}{100}*\frac{95}{100}\)
\(\frac{2}{25}*\frac{19}{20} = \frac{19}{250}*\frac{4}{4} = \frac{76}{1000}\) IS nBT
(1+2)
Using the above values for BT and nBT in the equation: (pBT)/(pBT + pnBT)
\(\frac{\frac{49}{1000}}{\frac{49}{1000}+\frac{76}{1000}}\)
\(\frac{\frac{49}{1000}}{\frac{125}{1000}}\)
\(\frac{49}{1000}*\frac{1000}{125}\)
\(\frac{49}{125}\) [multiply through by \(\frac{8}{8}\)]
\(\frac{392}{1000}\)
