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Thread: Probability of radar detection

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    Probability of radar detection




    Aircraft is in certain area, radar correctly registers its presence with prob. of 0.95. If it is not present, radar falsely registers its presence with prob. of 0.03. We assume that aircraft is present with prob of 0.1.

    So P(aircraft is present)= 0.1 and P(radar registers an aircraft) = 0.95 correct??
    Because I cannot figure out how to start on the questions below..

    Probability of false alarm
    Probability of missed detection
    Probability that an aircraft is present if not registered (conditional probability right?)

    guidance is appreciated!

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    Re: Probability of radar detection

    There are four possible states of this universe:
    • Aircraft is present, radar says aircraft is present
      Aircraft is present, radar says aircraft is not present
      Aricraft is not present, radar says aircraft is present
      Aircraft is not present, radar says aircraft is not present
    Start by computing the probability of each state. (Since these are all the possible states in this universe, their probabilities must sum to one. You can use that fact to check your work.)

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    Re: Probability of radar detection

    once i calculate those what does false alarm mean?

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    Re: Probability of radar detection

    A false alarm would be if you detect an aircraft but one isn't actually present.

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    Re: Probability of radar detection

    okay so P(radar detects aircraft | aircraft is not present), I got 0.1 probability...
    because equals the intersect of (0.1*0.95)/0.1 right? thats the law of conditional probability i was using..

    it also says for the last 2 questions the total probability law must be used which is what i think you stated in the last post

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    Re: Probability of radar detection

    I'm not sure what "the intersect of (0.1*0.95)/0.1" means.

    Look, you probably have some textbook or notes full of "law of this" and "law of that" stated in theorem-speak involving words like "intersect". Don't let it confuse you. You just need to keep two things in mind to figure this out.

    (1) To get the probability of A AND B happening, multiply their probabilities. For example: P(you have a dog) = 0.5, P(a dog has brown fur = 0.5), so P(you have a brown-haired dog) = 0.5*0.5=0.25)

    (2) To get the probability of A OR B happening, add their probabilities. For example: P(you have a brown eyes) = 0.5, P(you have blue eyes) = 0.3, so P(you have brown or blue eyes) = 0.8. In a unverse where all eyes are brown, blue or green, this implies that P(you have green eyes) = 0.2.

    I'll get you started by computing the first probability. P(an aircraft is present) = 0.1. P(a present aircraft is detected) = 0.95. So P(an aircraft is present and detected) = 0.1 * 0.95 = 0.095. Now you calculate the other three probabilities and get back to us.

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    Re: Probability of radar detection

    Quote Originally Posted by ichbin View Post
    (1) To get the probability of A AND B happening, multiply their probabilities.
    ...
    (2) To get the probability of A OR B happening, add their probabilities.
    assuming independence

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    Re: Probability of radar detection

    probability that aircraft is present and detected is 0.095?? that doesnt seem right but they math is definitely right.

    I got the probability of a false alarm to be 0.03 since P(aircraft present | radar registers aircraft) = (PA^B)/P(A)

    so that means probability of missed detection is P((radar registers aircraft)'|aircraft present) where ' means NOT... right?

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    Re: Probability of radar detection

    Why does 0.095 seem wrong to you? 0.095=9.5% of the time. An aircraft is only there 0.1=10% of the time, so the fraction of the time that it is there and detected is going to have to be smaller than 10%.

    How do you get 0.03 as the probability of a false alarm? You don't show your work.

    You seem to be getting ahead of yourself by trying to directly use Bayes' theorem without really being clear on what P(A|B) vs. P(B|A) vs. P(A) vs. P(B) mean here. It will all be much simpler and clearer if don't try to jump directly to the answer, but start with the small step I have asked you to take first: compute the overall probability of each of the four states.

    I computed one of them for you. Since every single on of them is the joint occurance of two events, every single one of them involves one single multiplication (no division!). To get the probability of plane there and detected, I multiplied the probability of plane there (0.1) and plane detected when there (0.95). To get the probability of plane not there and detected you will have to multiply the probability of plane not there (what is that?) and plane detected when not there (what is that?). Keep going and come back with those probabilities.

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    Re: Probability of radar detection

    P(aircraft not present)=0.9
    P(radar falsely registers when aircraft present)=1-0.95= 0.05 right?

    okay false alarm would be =(0.9)*(0.05) to get 0.045
    that seems right..

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    Re: Probability of radar detection

    Getting closer!

    Yes, P(aircraft not present) = 1 - 0.1 = 0.9! But the problem says P(detected when not present) = 0.03, so P(no aircraft but radar says there is) = 0.9 * 0.03 = 0.027. Now you have probabilities for two of the four states; just two more to go!

    Here's another hint: organize them into a table.

    \begin{matrix} & {\rm radar} + & {\rm radar} - \\
{\rm plane} + & 0.1 \times 0.95 & ? \\
{\rm plane} - & 0.9 \times 0.03 & ? \\
\end{matrix}

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    Re: Probability of radar detection

    so prob. of missed detection, is P(aircraft is there but radar does not register it)
    P(radar registers aircraft if present)=0.95 so
    P(radar doesnt register is present)=1-.95=0.05
    P(aircraft is present)=0.1

    P(missed detection)=(0.05)*0.1=0.005 !!!

    P(aircraft is not present and radar says not present)=0.9*0.05= 0.045 ??

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    Re: Probability of radar detection

    Very good. Yes, P(plane+ & radar-) = 0.1 * 0.05 = 0.005. But P(plane- & radar=) = 0.9 * 0.97 = 0.873. 0.97 is the complement of 0.03 for the plane not present case, just as 0.05 is the complement of 0.95 for the plane present case.

    So we now have the full table:

    \begin{matrix} & {\rm radar}+ & {\rm radar}- \\
{\rm plane}+ & 0.1 \times 0.95 = 0.095 & 0.1 \times 0.05 = 0.005 \\
{\rm plane}- & 0.9 \times 0.03 = 0.027 & 0.9 \times 0.97 = 0.873
\end{matrix}

    Note that the probabilities sum to 1, as they should. Take a moment to think about what this is telling you about the relative frequencies of all four cases. Just under 90% of the time there is no plane there and the alarm is not going off. The second most common situation, just under 10% of the time, is that a plane is there an the alarm is going off. These are the "working right" states. But just under 3% of the time there is no plane there and the alarm is going off anyway, and 1/2% of the time there is a plane there and the radar is missing it.

    Note that, if you sum across rows, you get 0.1 for the first row, which the the total probability that a plane is there, and 0.9 for the second row, which is the total probability that a plane is not there. But that's not so interesting, because we knew that when we started. What is more interesting is that, if you sum across columns, you get the total probabilities that the alarm is going off or not going off. That is new information. Go ahead and obtain those column totals.

    Once you have those totals, find this out: of all the times the alarm is not going off, in what fraction of those is a plane actually there? This is, by the way, the answer to your third part.

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    Re: Probability of radar detection

    so for the third part where its the probability that aircraft is present if it is not registered by the radar you would add 0.095+0.005 to get 0.1? or using conditional probability you would calculate it by 0.005/0.05 to get 0.1?
    im starting to doubt this is right...

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    Re: Probability of radar detection


    Follow the steps. What are the total column probabilities?

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