# Finding a missing airplane (Conditional probability? Confusing wording.)

• Sep 2nd 2011, 02:37 AM
Combinatus
Conditional probability question (with confusing wording)
LaTeX is giving me some serious parsing issues today, so I'll try to use it sparsely.

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Problem statement:

An airplane is reported as missing. Investigators assume that the airplane has disappeared in one of three given zones. Denote the probability that the airplane is found in zone k, if it is there, with 1- $a_k$. Different values of $a_k$ may depend on geographical reasons, differences in vegetation, et cetera.

Calculate the probability that the airplane is in zone k (k=1,2,3) if a search for the airplane in zone 1 didn't result in the airplane being found.

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First off, I'm not really sure how one failed attempt to look for an airplane would have an impact on where it actually is. Of course, if a large number of searches are being made for it in zone 1 without it being found, it would seem more likely for it to be in zone 2 or 3, but the more I think about it, the less sure I am that there's an explicit connection.

Assuming that there is, it may be sensible to assume an a priori probability of 1/3 for the airplane being in a given zone (before considering the effects of how $a_k$ influences the conditional probability). Then again, the wording "Investigators assume that the airplane has disappeared in one of three given zones" makes it sound like a bad idea to not take an "outside zones 1-3" region into account.

So, let $A_k$ (1 <= k <= 3) be the event that the airplane is in zone k. (We may also let $B$ be the event that the plane is not in either of the three zones.) Furthermote, let $F_j$ (1 <= j <= 3) be the event that the airplane is found in zone j.

What we're looking for, then, I think, is

P(A_1 | F_1^c), P(A_2 | F_1^c) and P(A_3 | F_1^c), respectively.

We get

P(A_1 | F_1^c) = (Bayes' theorem) = P(F_1^c | A_1)*P(A_1) / P(F_1^c) = $a_1$*P(A_1) / P(F_1^c) = $a_1$ / (3*P(F_1^c)).

The expressions for P(A_2 | F_1^c) and P(A_3 | F_1^c) don't simplify as nicely.

I've drawn probability trees and venn diagrams, but I haven't managed to get further than what's seen above. Any hints?
• Sep 3rd 2011, 02:36 AM
CaptainBlack
Re: Conditional probability question (with confusing wording)
Quote:

Originally Posted by Combinatus
LaTeX is giving me some serious parsing issues today, so I'll try to use it sparsely.

---

Problem statement:

An airplane is reported as missing. Investigators assume that the airplane has disappeared in one of three given zones. Denote the probability that the airplane is found in zone k, if it is there, with 1- $a_k$. Different values of $a_k$ may depend on geographical reasons, differences in vegetation, et cetera.

Calculate the probability that the airplane is in zone k (k=1,2,3) if a search for the airplane in zone 1 didn't result in the airplane being found.

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1. The looks do not have any effect on where the plane is, only on our assessment to the probability that the plane is in each of the possible locations. You may consider the problem embedded within a larger meta-problem where an ensemble of planes crash with the appropriate proportions of cases with the plane in each of the regions.

2. You do not need to worry about the possibility of the plane being elsewhere in this problem, one of the regions may be considered to include the elsewhere if there is such.

3. Let $p_1$, $p_2$ and $p_3$ denote the prior probabilities that the plane is in regions 1, 2 and 3 respectivly. A look is made in region 1 and the plane not found. Let $p_1^*$, $p_2^*$ and $p_2^*$ denote the posterior probabilities, then by Bayes theorem:

$p_1^*=\frac{p_1 a}{p_1a+(1-p_1)}$

Also $p_1^*+p_2^*+p_3^*=1$ and $p_2^*/p_3^*=p_2/p_3$ which will allow the computation of $p_2^*$ and $p_3^*$

CB
• Sep 3rd 2011, 01:43 PM
Combinatus
Re: Conditional probability question (with confusing wording)
Thanks for your response! I hadn't heard of posterior probabilities before, but reading a bit on Wikipedia makes it apparent that it's what the problem is asking for. Still, I'm confused about something, and would like to ask a dumb question about the part quoted below.

Quote:

Originally Posted by CaptainBlack
[...]by Bayes theorem:

$p_1^*=p_1a_1$

That looks a lot like my P("Plane is in region 1 after failing to find it in region 1") = $P(A_1 | F_1^c) = a_1 P(A_1) / P(F_1^c)$ statement. If it is equivalent, then $P(F_1^c)$ should be 1. But why? Perhaps it makes sense to think of it as "the probability of of not finding the plane in region 1, before looking for it"? Or, am I missing something completely?

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Edit: Great, I'm even more confused now. I had assumed that the LHS of $\frac{p_2^*}{p_3^*} = \frac{p_2}{p_3}$ was equivalent to, in my symbols, $\frac{P(A_2|F_1^c)}{P(A_3|F_1^c)} = \frac{P(F_1^c|A_2)\cdot P(A_2)}{P(F_1^c|A_3)\cdot P(A_3)}$, where $P(F_1^c|A_2)$ and $P(F_1^c|A_3)$ would both be equal to 1, as "the probability of not finding the plane in region 1 if it is in region 2 (or 3)" is 1, thus giving me the equivalent of your RHS.

However, if that was so, then $p_2^* = P(A_2|F_1^c) = \frac{P(F_1^c|A_2) \cdot P(A_2)}{P(F_1^c)} = \frac{1 \cdot P(A_2)}{1} = p_2$, which doesn't make any sense. :(
• Sep 3rd 2011, 10:11 PM
CaptainBlack
Re: Conditional probability question (with confusing wording)
Quote:

Originally Posted by Combinatus
Thanks for your response! I hadn't heard of posterior probabilities before, but reading a bit on Wikipedia makes it apparent that it's what the problem is asking for. Still, I'm confused about something, and would like to ask a dumb question about the part quoted below.

That looks a lot like my P("Plane is in region 1 after failing to find it in region 1") = $P(A_1 | F_1^c) = a_1 P(A_1) / P(F_1^c)$ statement. If it is equivalent, then $P(F_1^c)$ should be 1. But why? Perhaps it makes sense to think of it as "the probability of of not finding the plane in region 1, before looking for it"? Or, am I missing something completely?

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(

There is a term missing in my earlier post:

$p_1^*=\frac{p_1 a}{p_1a+(1-p_1)}$

That is I forgot to divide by the probability that the plane is not found when we look in region 1, which is the sum of the probabilities of it being in region 1 and not seen plus the probability that it is not in region 1.

CB
• Sep 3rd 2011, 10:22 PM
CaptainBlack
Re: Conditional probability question (with confusing wording)
Quote:

Originally Posted by Combinatus

Edit: Great, I'm even more confused now. I had assumed that the LHS of $\frac{p_2^*}{p_3^*} = \frac{p_2}{p_3}$ was equivalent to, in my symbols, $\frac{P(A_2|F_1^c)}{P(A_3|F_1^c)} = \frac{P(F_1^c|A_2)\cdot P(A_2)}{P(F_1^c|A_3)\cdot P(A_3)}$, where $P(F_1^c|A_2)$ and $P(F_1^c|A_3)$ would both be equal to 1, as "the probability of not finding the plane in region 1 if it is in region 2 (or 3)" is 1, thus giving me the equivalent of your RHS.

However, if that was so, then $p_2^* = P(A_2|F_1^c) = \frac{P(F_1^c|A_2) \cdot P(A_2)}{P(F_1^c)} = \frac{1 \cdot P(A_2)}{1} = p_2$, which doesn't make any sense. :(

Your $P(F_c^1)=P(\text{not seen in r1})=p_1a+(1-p_1)\ne 1$.

Now:

$p_2^*=\frac{P({\text{not seen in r1}}|{\text{in r2}})p_2}{P(\text{not seen in r1})}$

but: $P({\text{not seen in r1}}|{\text{in r2}})=1$.

So:

$p_2^*=\frac{p_2}{P(\text{not seen in r1})}$

and similarly:

$p_3^*=\frac{p_3}{P(\text{not seen in r1})}$

So:

$\frac{p_2^*}{p_3^*}=\frac{p_2}{p_3}$

CB
• Sep 4th 2011, 07:09 AM
Combinatus
Re: Conditional probability question (with confusing wording)
Quote:

Originally Posted by CaptainBlack
Your $P(F_c^1)=P(\text{not seen in r1})=p_1a+(1-p_1)\ne 1$.

Wonderful! I wish I would had thought of that right away. Fortunately, I learnt quite a bit about prior and posterior probabilities from this thread, so starting it was absolutely worth it. Thank you!