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A small, experimental plane has three engines, one of which
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Updated on: 16 Sep 2013, 00:48
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A small, experimental plane has three engines, one of which is redundant. That is, as long as two of the engines are working, the plane will stay in the air. Over the course of a typical flight, there is a 1/3 chance that engine one will fail. There is a 75% probability that engine two will work. The third engine works only half the time. What is the probability that the plane will crash in any given flight? (A) 7/12 (B) 1/4 (C) 1/2 (D) 7/24 (E) 17/24
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Originally posted by Onell on 15 Mar 2011, 22:42.
Last edited by Bunuel on 16 Sep 2013, 00:48, edited 1 time in total.
Renamed the topic, edited the question and added the OA.




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Re: xperimental plane
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16 Mar 2011, 00:01
Onell wrote: A small, experimental plane has three engines, one of which is redundant. That is, as long as two of the engines are working, the plane will stay in the air. Over the course of a typical flight, there is a 1/3 chance that engine one will fail. There is a 75% probability that engine two will work. The third engine works only half the time. What is the probability that the plane will crash in any given flight?
(A) 7/12 (B) 1/4 (C) 1/2 (D) 7/24 (E) 17/24 Lets see the probability of failure of each engine Engine 1  1/3 (hence , non failure probability = 2/3) Engine 2  1/4 (hence , non failure probability = 3/4) Engine 3  1/2 (hence , non failure probability = 1/2) Plane will fail in two scenarios  2 engines fail or all three fail 2 engines can fail in three scenarios wnere 1,2 or 3 dont fail but the other two fail. So, total proabability of plane failure = (1/3*1/4*1/2) + (1/3*1/4*1/2+2/3*1/4*1/2+3/4*1/3*1/2) = 7/24. Answer D




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Re: xperimental plane
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22 Apr 2011, 12:14
Nice question The probability works out like this, if there are three engines namely A,B,C & plane will crash if 2 or more engines will fail thus the probability of plane getting crash is f(A)*f(B)*f(C) + A*f(B)*f(C) + B*f(A)*f(C) + C*f(B)*f(A) where f(A), f(B), f(C) are probability of respective engines failing.. Onell wrote: A small, experimental plane has three engines, one of which is redundant. That is, as long as two of the engines are working, the plane will stay in the air. Over the course of a typical flight, there is a 1/3 chance that engine one will fail. There is a 75% probability that engine two will work. The third engine works only half the time. What is the probability that the plane will crash in any given flight?
(A) 7/12 (B) 1/4 (C) 1/2 (D) 7/24 (E) 17/24
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Re: xperimental plane
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16 Sep 2013, 00:14
CASE ONE: Engine 1 fails, Engine 2 fails, Engine 3 works = 1/3* 1/4 * 1/2 = 1/24
CASE TWO: Engine 1 fails, Engine 2 works, Engine 3 fails = 1/3*3/4*1/2= 3/24
CASE THREE: Engine 1 works, Engine 2 fails, Engine 3 fails = 2/3*1/4*1/2=2/24
CASE FOUR: Engine 1 fails, Engine 2 fails, Engine 3 fails = 1/4*1/4*1/2=1/24
Adding, 1/24+3/24+2/24+1/24 = 7/24
Ans : D



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Re: A small, experimental plane has three engines, one of which
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07 May 2014, 09:15
I just took the first 3 cases and ended up choosing 1/4 How does it matter that all 3 engine fail 1. Fails 2. Fails 3.Works Result  Crash 1 Fails 2 Fails 3 Fails Result  Crash So after any of the 2 fail..why even bother to consider the third one?
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Re: xperimental plane
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16 Mar 2011, 00:02
beyondgmatscore wrote: Onell wrote: A small, experimental plane has three engines, one of which is redundant. That is, as long as two of the engines are working, the plane will stay in the air. Over the course of a typical flight, there is a 1/3 chance that engine one will fail. There is a 75% probability that engine two will work. The third engine works only half the time. What is the probability that the plane will crash in any given flight?
(A) 7/12 (B) 1/4 (C) 1/2 (D) 7/24 (E) 17/24 Lets see the probability of failure of each engine Engine 1  1/3 (hence , non failure probability = 2/3) Engine 2  1/4 (hence , non failure probability = 3/4) Engine 3  1/2 (hence , non failure probability = 1/2) Plane will fail in two scenarios  2 engines fail or all three fail 2 engines can fail in three scenarios wnere 1,2 or 3 dont fail but the other two fail. So, total proabability of plane failure = (1/3*1/4*1/2) + (1/3*1/4*1/2+2/3*1/4*1/2+3/4*1/3*1/2) = 7/24. Answer D awesome thanks for your help



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Re: xperimental plane
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16 Mar 2011, 19:29
I got 1/24 missed one of the cases
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Re: xperimental plane
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17 Mar 2011, 02:27
The plane will crash if 2 out of 3 crash or 3/3 crash 1  C 2  C 3  ok  1/3 * 1/4 * 1/2 1  C 2  ok 3  C  1/3 * 3/4 * 1/2 1  ok 2  C 3  C  2/3 * 1/4 * 1/2 1  C 2  C 3  C  1/3 * 1/4 *1/2 = 1/24 + 1/24 + 3/24 + 2/24 = 7/24
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Re: xperimental plane
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11 Apr 2011, 21:02
GMAT could have easily tricked us by adding one of the answer options as 1/24.... Even I missed the second case... :'(



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Re: xperimental plane
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12 Apr 2011, 00:02
In probability questions the trap answer is just the multiple of the numbers in the question. i.e. if you multiply 1/3 * 1/4 * 1/2 = 1/24 is trap answer The other trap answer could be 2/3 * 3/4 * 1/2 = 6/24 is trap answer So lets say you have 30 secs and you want to guess the answer then B, C are ruled out because they can be traps. You best guess is A, D, E. So you have 33% chances of being correct. (A) 7/12 (B) 1/4 (C) 1/2 (D) 7/24 (E) 17/24 pesfunk wrote: GMAT could have easily tricked us by adding one of the answer options as 1/24.... Even I missed the second case... :'(



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Re: xperimental plane
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30 Apr 2011, 22:27
A! = engine A does not work. (A*B!*C!) + (A!*B!*C) + (A!*B*C!) + (A!*B!*C!)
giving 7/24.



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Re: A small, experimental plane has three engines, one of which
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08 May 2014, 03:31
JusTLucK04 wrote: I just took the first 3 cases and ended up choosing 1/4
How does it matter that all 3 engine fail
1. Fails 2. Fails 3.Works Result  Crash
1 Fails 2 Fails 3 Fails Result  Crash
So after any of the 2 fail..why even bother to consider the third one? There is a possibility that all 3 will fail simultaneously, hence you have to consider this case too.
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Re: A small, experimental plane has three engines, one of which
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09 May 2014, 02:11
Bunuel wrote: JusTLucK04 wrote: I just took the first 3 cases and ended up choosing 1/4
How does it matter that all 3 engine fail
1. Fails 2. Fails 3.Works Result  Crash
1 Fails 2 Fails 3 Fails Result  Crash
So after any of the 2 fail..why even bother to consider the third one? There is a possibility that all 3 will fail simultaneously, hence you have to consider this case too. What was I thinking...
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Re: A small, experimental plane has three engines, one of which
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01 Nov 2016, 05:15
Bunuel wrote: JusTLucK04 wrote: I just took the first 3 cases and ended up choosing 1/4
How does it matter that all 3 engine fail
1. Fails 2. Fails 3.Works Result  Crash
1 Fails 2 Fails 3 Fails Result  Crash
So after any of the 2 fail..why even bother to consider the third one? There is a possibility that all 3 will fail simultaneously, hence you have to consider this case too. Banuel, I did it by subtracting (all the possibilities when plane does not crash) from 1. Is this a wrong approach? E1 E2 E3 be Engines and F  Fail NF  No Fail :. 1  [ (E1F * E2NF * E3NF) + (E1NF * E2F * E3NF) + (E1NF * E2NF * E3F) ]WHERE E1NF = 2/3 & E1F = 1/3 E2NF= 75/100 & E2F = 25/100 E3NF= 1/2 & E3F = 1/2I DID NOT get the right answer.



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Re: A small, experimental plane has three engines, one of which
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28 Jun 2017, 22:55
deekshant111 wrote: Bunuel wrote: JusTLucK04 wrote: I just took the first 3 cases and ended up choosing 1/4
How does it matter that all 3 engine fail
1. Fails 2. Fails 3.Works Result  Crash
1 Fails 2 Fails 3 Fails Result  Crash
So after any of the 2 fail..why even bother to consider the third one? There is a possibility that all 3 will fail simultaneously, hence you have to consider this case too. Banuel, I did it by subtracting (all the possibilities when plane does not crash) from 1. Is this a wrong approach? E1 E2 E3 be Engines and F  Fail NF  No Fail :. 1  [ (E1F * E2NF * E3NF) + (E1NF * E2F * E3NF) + (E1NF * E2NF * E3F) ]WHERE E1NF = 2/3 & E1F = 1/3 E2NF= 75/100 & E2F = 25/100 E3NF= 1/2 & E3F = 1/2I DID NOT get the right answer. Hello Mr.Deekshant111: You have missed out a case of all 3 engines working properly. So the final answer would be 7/24. 1 [ 1/3x3/4x1/2 + 2/3x1/4x1/2 + 2/3 X3/4 X1/2 + 2/3 x 3/4 X1/2 ] So the approach is right. Sent from my ONE A2003 using GMAT Club Forum mobile app



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Re: A small, experimental plane has three engines, one of which
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29 Jun 2017, 00:11
Onell wrote: A small, experimental plane has three engines, one of which is redundant. That is, as long as two of the engines are working, the plane will stay in the air. Over the course of a typical flight, there is a 1/3 chance that engine one will fail. There is a 75% probability that engine two will work. The third engine works only half the time. What is the probability that the plane will crash in any given flight?
(A) 7/12 (B) 1/4 (C) 1/2 (D) 7/24 (E) 17/24 1. Let engine 1 fail. Then at least one of the other two should fail. Probability is 1/3* (1  3/4*1/2)=5/24 2. Let engine 1 work and the other two fail. Probability is 2/3*(1/4*1/2) =2/24 3. Final probability is 5/24+2/24=7/24
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Re: A small, experimental plane has three engines, one of which
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29 Jun 2017, 00:54
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Re: A small, experimental plane has three engines, one of which
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31 Aug 2019, 03:16
can this be solved via solved by combination ? 3c2 (both engine fail) and 3c3 all three fail..but i am not sure how to proceed further in these 2 cases..




Re: A small, experimental plane has three engines, one of which
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