first of all I want to say hey to all! I have been a long time lurker and follower of these forums and have been known to find a lot of your answers helping me out throughout my college life.

I also know that you just don't like to spit out the answers for those who don't put in the effort so here it goes:

This is a very basic conditional probability question, but I can't seem to find a trick in solving these questions and they always get me. Any help would be greatly appreciated. The question is as follows:

The air rescue service of the armed forces divided the flight path of a downed plane into search sectors. In one mission a search plane will overfly every square kilometer of the designated sector. From past experience, one knows however that in one mission there is only a 50% chance of spotting a plane down in a sector of tundra. There is a 40% chance of spotting a plane down in a forested sector and a 15% chance of spotting a plane down in a lake sector. (a) A small plane is down in an area containing two sectors of tundra, one lake sector and one forested sector and a priori the plane has an equal chance of being down in any one of the fours sectors. The mission director decides to use the first mission to search the two sectors of tundra. What is the probability of spotting the downed plane?

(b) Given that we did not find the plane in the tundra sections, what is the probability that the plane is (i) in the tundra sections? (ii) in the forested section? (iii) in the lake sector?

Ok, so for a) what I did was basically the plane is going over the two sectors of Tundra which has a probability of 50% of containing the downed plane. So P(T) . P(T) would be 0.5*0.5 = 0.25 is the probability of finding the plane in the two sectors of Tundra, right?

for b) (i) I said P(Finding Plane | Tundra) = P(Finding Plane intersect Tundra) / P(Finding Plane) (ii) P(Finding Plane | Forest) = P(Finding Plane intersect Forest) / P(Finding Plane) (iii) P(Finding Plane | Lake) = P(Finding Plane intersect Lake) / P(Finding Plane)

P(Finding Plane) would be the total probability of finding the plane in each sector so given that it has an equal chance of being in either sector 2/4 * P(Tundra) + 1/4 * P(Forest) + 1/4*P(Lake)

I am not sure about my answers nor my approach to this problem. Any tricks or tips to solve this problem or conditional probability in general?

  • $\begingroup$ Note that P(Finding Plane | Tundra) = P(Finding Plane intersect Tundra) / P(Tundra) not P(Finding Plane | Tundra) = P(Finding Plane intersect Tundra) / P(Finding Plane) $\endgroup$ – E.O. Jan 29 '12 at 21:47
  • $\begingroup$ any tricks on solving such questions? what should I look for first? $\endgroup$ – Alistair Jan 29 '12 at 23:26

If you are finding this kind of problem difficult or just want to check your results, you could try this technique.

Suppose this happened $400$ times (you don't have to use this number, but there are four equally likely sectors and you are dealing with percentages). Then the plane would be in the tundra an expected $200$ times, in the lake $100$ times and in the forest $100$ times. It would be found in the tundra an expected $100$ times. $\dfrac{100}{400} = 0.25$ so you have the right result for (a).

So exclude the expected $100$ times it is found in the tundra on the first mission, leaving $300$, of which $100$ are in the tundra, $100$ in the lake and $100$ in the forest. So for (b) each of these has a $\dfrac{100}{300} \approx 0.333$ probability of being true.

If you want to do this formally then for (a) you have

$\Pr(\text{found in tundra|down in tundra})\Pr(\text{down in tundra}) = \dfrac{50}{100} \times \dfrac{2}{4} = \dfrac{1}{4}.$

For (b), one of the calculations is

$\Pr(\text{down in lake|not found in tundra}) =\dfrac{\Pr(\text{down in lake})\Pr(\text{not found in tundra|down in lake})}{\Pr(\text{not found in tundra})} $ $= \dfrac{\frac{1}{4} \times 1}{1 - \frac{1}{4}} = \dfrac{1}{3}.$

  • $\begingroup$ Thanks, let me try this question out with your suggestions. How can I upvote or give you reputation without having any of my own? : ) $\endgroup$ – Alistair Jan 29 '12 at 23:56
  • $\begingroup$ @Alistair You can accept the answer, if you think it is the best or only one which meets your needs. $\endgroup$ – Henry Jan 30 '12 at 0:01
  • $\begingroup$ I understood perfectly the formal calculation part, however the first part I am having a bit of trouble grasping. We have 2 sections of Tundra, therefore we expect the plane to be in the Tundra sections 200 times. So why do we say it will be found 100/400 times for part a? Also for part b) why do we exclude the expected 100 times it is found in the tundra? Because we are dealing with 2 tundra sections? Thanks for the help again. $\endgroup$ – Alistair Jan 30 '12 at 0:46
  • $\begingroup$ @Alistair: it goes down in the tundra an expected 200 times and is found half of these times, i.e. an expected 100 times, out of the total 400 times. $\endgroup$ – Henry Jan 30 '12 at 0:56

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