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 A specific kind of small airplane used for dusting
crops has two identical engines functioning independently  
of  each  other,  but  can  be  flown  with only one engine. 
The past records of this kind of engine indicate a 3% mid-air 
failure rate for each engine. One  such  plane  crashed  while  dusting
crops, and mid-air engine failure was suspected.
What is the probability that such an airplane is
likely to crash due to mid-air engine failure?

The way to solve this problem is by using conditional probability. Say A is the probability of crashing, and F is the probability of engine failure. Let F1 the probability of getting failure (3%) and F2 the probability of getting failure also on other engine. What A should be? what is the best way to solve this problem? Thanks

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2 Answers 2

up vote 2 down vote accepted

This seems a rather i'll stated problem. If you interpret "such an airplane" as an airplane that crashed, then you are finding the probability that both engines failed given that it crashed. So, with your notation, you want $P(F_1\cap F_2 | A ) ={P(F_1\cap F_2\cap A)\over P(A) }$. You can simplify and calculate $P(F_1\cap F_2\cap A)=P(F_1\cap F_2 )=(.03)^2$. However, nothing in the problem allows you to calculate $P(A)$. You're at a dead end.

Perhaps, the problem meant for you to find the probability $P(F_1\cap F_2)$.

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Whats the difference of this P(F1∩F2∩A) with this: P(F1∩F2) P(A|F1∩F2). Can i use it and everything will be fine? –  IvanMatala Jan 26 '12 at 13:57
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@blackandyellow You can use that, but note $P(A|F_1\cap F_2)=1$... –  David Mitra Jan 26 '12 at 14:20
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I am not sure if you need conditional probability for doing this problem. I'd think of it this way. Let $F_i$ be that the $i^{th}$ engine fails in mid-air for $i=1,2$. So, the plane would crash if both the engines failed simultaneously. And, this event is represented as $F_1 \cap F_2$.

So, $$P(F_1 \cap F_2)=P(F_1) \cdot P(F_2)~~~~ \text{since $F_1$ and $F_2$ are independent}$$

Now, $P(F_1)=P(F_2)=\frac{3}{100}$ gives you, that, the required probability is $\dfrac{9}{10^4}$

Hope this helps.

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nice explanation –  IvanMatala Jan 26 '12 at 13:26
    
@user963499 I have solved an ideal case without having to bump into dead ends such as the one by David Mitra here! So, I suggest you go through that. Because, problems in probability are subtly worded, that you might miss something. Thank you for your compliment anyway. –  user21436 Jan 26 '12 at 13:29
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