probability that the plane will crash

[abhi332]

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
[spoiler]
OA:D[/spoiler]

[harsh.champ]

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
[spoiler]
OA:D[/spoiler]

We have to consider 2 cases:-1)When 2 engines have failed
2)When all 3 engines have failed
Case 1:-Probability that the plane will crash = 1/3 x 1/4 x1/2 + 2/3 x 1/4 x 1/2 + 1/3 x 3/4 x 1/2 = 7/24

Hey abhi,
I wanted to ask why don't we consider the Case 2??

[abhi332]

you don't have consider 2nd one because if 2 engine fails than plane will surely crash.

No need to consider when all engine fails.

[abhi332]

Hey Harsh, I am not sure..

[abhi332]

hey harsh,

Are you sure calculation is correct

Probability that the plane will crash = 1/3 x 1/4 x1/2 + 2/3 x 1/4 x 1/2 + 1/3 x 3/4 x 1/2

I am getting 0.25

but the ans is 0.29

[harsh.champ]

Hey Harsh, I am not sure..

I guess we need help from some expert over here.
Whats the source??

[harsh.champ]

you don't have consider 2nd one because if 2 engine fails than plane will surely crash.

No need to consider when all engine fails.

But still,that would be a different case,
so why should we leave it??
Even if all 3 fails ,still the plane will crash.
According to me,the OA is wrong or the question framed very poorly...
Whatever be the case,I dont think such ambiguous questions can appear on test-day.
Wat-say??

[harsh.champ]

hey harsh,

Are you sure calculation is correct

Probability that the plane will crash = 1/3 x 1/4 x1/2 + 2/3 x 1/4 x 1/2 + 1/3 x 3/4 x 1/2

I am getting 0.25

but the ans is 0.29

Dr. =24
Nr. = 1+2+3 =6 :!:

Hey,we have to consider the 3rd case over here.
Got it.
Chalo, atleast my concept was right,if not the calculation :mrgreen:

[abhi332]

Following cases were considered while solving this question in the actual explanation

CASE ONE: Engine 1 fails, Engine 2 fails, Engine 3 works
CASE TWO: Engine 1 fails, Engine 2 works, Engine 3 fails
CASE THREE: Engine 1 works, Engine 2 fails, Engine 3 fails
CASE FOUR: Engine 1 fails, Engine 2 fails, Engine 3 fails

[harsh.champ]

Following cases were considered while solving this question in the actual explanation

CASE ONE: Engine 1 fails, Engine 2 fails, Engine 3 works
CASE TWO: Engine 1 fails, Engine 2 works, Engine 3 fails
CASE THREE: Engine 1 works, Engine 2 fails, Engine 3 fails
CASE FOUR: Engine 1 fails, Engine 2 fails, Engine 3 fails

Look at the 4th case.
It's the same as mine.
I guess we do have to take the 4th case over here.

[Testluv]

In these scenario or case probability questions, you want to think hard about the different ways the desirable event can occur.

What's the desirable event here?
...the plane crashing.

How can the plane crash?

If at least 2 engines fail...so if 2 or 3 engines fail.

How can 2 engines fail?

If the first and second fail OR if the first and third fail OR if the second and third fail.

That's three cases. And of course there's only one case for all 3 engines failling...that's a total of four cases.

The reason we have to consider this fourth case is that the three engines might fail simultaneously rather than sequentially...in other words, we can't just assume that the engines fail in sequence. If all three engines fail at the same time, then that is a distinct case from where only two of the engines are failing.

[abhi332]

Thanks Testluv, I understood this now clearly

[Scott@TargetTestPrep]

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

We are given that a plane will stay in the air as long as at least two of the three engines are working. So, to determine the probability that the plane will crash, we need to determine the probability that at least two of the three engines are not working.

The probability of engine one not working is 1/3, the probability of engine two is not working is 1/4, and the probability of engine 3 not working is 1/2.

Scenario 1:

1 does not work, 2 does not work, 3 works

1/3 x 1/4 x 1/2 = 1/24

Scenario 2:

1 does not work, 2 works, 3 does not work

1/3 x 3/4 x 1/2 = 3/24

Scenario 3:

1 works, 2 does not work, 3 does not work

2/3 x 1/4 x 1/2 = 2/24

Scenario 4:

1 does not work, 2 does not work, 3 does not work

1/3 x 1/4 x 1/2 = 1/24

Thus, the probability that the plane will crash is 1/24 + 3/24 + 2/24 + 1/24 = 7/24.

Answer: D