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# An airport screens bags for forbidden items, and an alarm is supposed

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Senior Manager
Joined: 13 Jan 2018
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Location: India
Concentration: Operations, General Management
GMAT 1: 580 Q47 V23
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WE: Consulting (Consulting)
An airport screens bags for forbidden items, and an alarm is supposed  [#permalink]

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14 Apr 2019, 02:52
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Difficulty:

75% (hard)

Question Stats:

33% (02:09) correct 67% (01:51) wrong based on 27 sessions

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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?

A) If a bag contains a forbidden item, there is a 98% chance that it triggers the alarm.

B) If a bag doesn't contain a forbidden item, there is an 8% chance that it triggers the alarm.

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Chaitanya

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Manager
Joined: 23 Aug 2017
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Re: An airport screens bags for forbidden items, and an alarm is supposed  [#permalink]

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14 Apr 2019, 07:06
Pl provide the solution also.
Math Expert
Joined: 02 Aug 2009
Posts: 7981
Re: An airport screens bags for forbidden items, and an alarm is supposed  [#permalink]

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14 Apr 2019, 08:19
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eswarchethu135 wrote:
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?

A) If a bag contains a forbidden item, there is a 98% chance that it triggers the alarm.

B) If a bag doesn't contain a forbidden item, there is an 8% chance that it triggers the alarm.

So the alarm bell has been triggered..
Now there are two cases - genuine ( let the prob be g) and false ( let the probability be f)
So our answer will be based on two cases that the bell has already triggered..
Incase of genuine cases - (5% forbidden )*g
Incase of false cases -(95% clean baggage)*f
Probability = (5% forbidden )*g/((5% forbidden )*g +(95% clean baggage)*f)

Thus we are looking for value of g and f..
Statement I tells us g is 0.98 and statement II tells us that f is 0.08..
Thus probability =0.05*0.98/( 0.05*0.98+0.08*0.95)

C
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Re: An airport screens bags for forbidden items, and an alarm is supposed   [#permalink] 14 Apr 2019, 08:19
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# An airport screens bags for forbidden items, and an alarm is supposed

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