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-$\displaystyle a_k$. Different values of $\displaystyle a_k$ 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 $\displaystyle a_k$ 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 $\displaystyle A_k$ (1 <= k <= 3) be the event that the airplane is in zone k. (We may also let $\displaystyle B$ be the event that the plane is not in either of the three zones.) Furthermote, let $\displaystyle F_j$ (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) = $\displaystyle a_1$*P(A_1) / P(F_1^c) = $\displaystyle a_1$ / (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?

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.

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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-$\displaystyle a_k$. Different values of $\displaystyle a_k$ 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 $\displaystyle p_1$, $\displaystyle p_2$ and $\displaystyle p_3$ 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 $\displaystyle p_1^*$, $\displaystyle p_2^*$ and $\displaystyle p_2^*$ denote the posterior probabilities, then by Bayes theorem:

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

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

CB

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:

$\displaystyle p_1^*=p_1a_1$

That looks a lot like my P("Plane is in region 1 after failing to find it in region 1") = $\displaystyle P(A_1 | F_1^c) = a_1 P(A_1) / P(F_1^c)$ statement. If it is equivalent, then $\displaystyle P(F_1^c)$ 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 $\displaystyle \frac{p_2^*}{p_3^*} = \frac{p_2}{p_3}$ was equivalent to, in my symbols, $\displaystyle \frac{P(A_2|F_1^c)}{P(A_3|F_1^c)} = \frac{P(F_1^c|A_2)\cdot P(A_2)}{P(F_1^c|A_3)\cdot P(A_3)}$, where $\displaystyle P(F_1^c|A_2)$ and $\displaystyle P(F_1^c|A_3)$ 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 $\displaystyle p_2^* = P(A_2|F_1^c) = \frac{P(F_1^c|A_2) \cdot P(A_2)}{P(F_1^c)} = \frac{1 \cdot P(A_2)}{1} = p_2$, which doesn't make any sense. :(

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") = $\displaystyle P(A_1 | F_1^c) = a_1 P(A_1) / P(F_1^c)$ statement. If it is equivalent, then $\displaystyle P(F_1^c)$ 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:

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

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

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 $\displaystyle \frac{p_2^*}{p_3^*} = \frac{p_2}{p_3}$ was equivalent to, in my symbols, $\displaystyle \frac{P(A_2|F_1^c)}{P(A_3|F_1^c)} = \frac{P(F_1^c|A_2)\cdot P(A_2)}{P(F_1^c|A_3)\cdot P(A_3)}$, where $\displaystyle P(F_1^c|A_2)$ and $\displaystyle P(F_1^c|A_3)$ 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 $\displaystyle p_2^* = P(A_2|F_1^c) = \frac{P(F_1^c|A_2) \cdot P(A_2)}{P(F_1^c)} = \frac{1 \cdot P(A_2)}{1} = p_2$, which doesn't make any sense. :(

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

Now:

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

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

So:

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

and similarly:

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

So:

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

CB

Re: Conditional probability question (with confusing wording)

Quote:

Originally Posted by

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

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!