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

Question

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

(A) 7/12
(B) 1/4
(C) 1/2
(D) 7/24
(E) 17/24

beyondgmatscore · Accepted Answer

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

ARUNPLDb · Answer

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

subhashghosh · Answer

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

MSDHONI · Answer

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..

JusTLucK04 · Answer

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?

Bunuel · Answer

There is a possibility that all 3 will fail simultaneously, hence you have to consider this case too.

JusTLucK04 · Answer

What was I thinking... :oops:

deekshant111 · Answer

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 -

WHERE 
 E1NF = 2/3 & E1F = 1/3
E2NF= 75/100 & E2F = 25/100
E3NF= 1/2 & E3F = 1/2

I DID NOT get the right answer.

GouthamNandu · 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-

So the approach is right.

Sent from my ONE A2003 using  GMAT Club Forum mobile app

SVaidyaraman · Answer

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

MathRevolution · Answer

P(Engine_1 Works): \frac{2 }{ 3} -------------------P(Engine_1 fails): \frac{1 }{ 3}

P(Engine_2 Works): \frac{3 }{ 4} -------------------P(Engine_2 fails): \frac{1 }{ 4}

P(Engine_3 Works): \frac{1 }{ 2} -------------------P(Engine_3 fails): \frac{1 }{ 2}

The plane stays in the air even if two engines are working.

The probability that the plane will crash in any given flight: 1 -

All three works:  \frac{2}{3} * \frac{3}{4} * \frac{1}{2} = \frac{1}{4}

2 works and 1 fails:  E_1 * E_2 * E_3' + E_1' * E_2 * E_3 + E_1 * E_2' * E_3

=>  \frac{1}{4} + \frac{1}{8} + \frac{1}{12}

=> 1 -  (\frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{12})

=> 1 -  \frac{17}{24}

=>  \frac{7}{24}

Answer D

shivamasthana · Answer

Why are we not considering the statement given in the question that one of the engines is already redundant? We might have to consider cases in which 1 engine is already not in use and then take out the probability. Would be helpful if someone can explain this.

Bunuel · Answer

I think you are misinterpreting the question. "One of the three engines is redundant" means that the plane, while equipped with three engines, can still fly with any two of them as long as they are functional. We have been given the probabilities of each engine failing and are asked the probability that the plane will crash in any given flight, which implies we need to find the probability that at least two of the three engines fail.

Hope it helps.

Kinshook · Answer

Given: 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.

Asked: What is the probability that the plane will crash in any given flight?

In case, no engine is working or 1 out of 3 engines is working, the plane will crash.

Engine 1: 
Probability of Engine 1 failing = 1/3
Probability of Engine 1 working = 1 -1/3 = 2/3

Engine 2: 
Probability of Engine 2 working = 75% = 3/4
Probability of Engine 2 failing = 100% - 75% = 25% = 1/4

Engine 3: 
Probability of Engine 3 working = 1/2
Probability of Engine 3 failing = 1 - 1/2 = 1/2

The probability of plane crasing = The probability of all engines failing + The probability of only 1 engine working
= 1/3*1/4*1/2 + 2/3*1/4*1/2 + 1/3*3/4*1/2 + 1/3*1/4*1/2 = 1/24 + 2/24 + 3/24 + 1/24 = 7/24

IMO D

OokandGluk · Answer

Bunuel  ,

Why do we even consider a third engine when two engines have already failed (which leads to plane crash)?

Regor60 · Answer

Multiplication works by weighting each factor against the other.

Your approach would be to assume the probability of a 3rd engine "event" is 1, since there's no reason to discriminate the event into fail or not fail.

The problem with that approach is that the question REQUIRES using probabilities for fail and not fail and there's no way of disentangling.

The other way to look at it is to say the probability of a crash is:

1 - probability of one or no engines failing

Posted from my mobile device

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