Probability mass function

Discussion in 'Physics & Math' started by kingwinner, Oct 1, 2008.

  1. kingwinner Registered Senior Member

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    796
    Consider a system of water flowing through valves from A to B as shown in the diagram. Valves 1, 2, and 3 operate independently, and each correctly opens on signal with probability 0.8. Find the probability distribution / probability mass function for Y, the number of open paths from A to B after the signal is given (Note that Y can take on the values 0, 1, and 2.)

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    Note: Probability mass function just means finding the probabilities of all possible outcomes.

    I don't understand the question itself so I can't get anywhere (coloured in red). Does anyone actually understand what is going on in this scenario? I would truly appreciate if somebody can explain it to me.
     
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  3. Vkothii Banned Banned

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    It looks easy; but I might not know what I'm talking about here.

    Is the implication that there are three possible valve 'states'; i.e. all closed, (equivalent to either 2 or 3 open); 1 open (equivalent to 1 and 3, or 1 and 2 open); 1 2 and 3 open? Try writing a truth table for the valve states, maybe?
     
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  5. kingwinner Registered Senior Member

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    796
    Thanks, but do you have any idea what is going on with the "signal" that the problem talks about twice?
     
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  7. Vkothii Banned Banned

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    3,674
    It just means a valve can be open or closed.
    With three valves, that's 2^3 or eight possibilities, but as I outlined that gives three possible 'states' for the circuit.

    Let's see; a table for each valve state (on or off), and the resulting output goes like this:

    signal | B (output)
    0 0 0 | 0
    0 0 1 | 0
    0 1 0 | 0
    0 1 1 | 0.8 x 0.8 (?)
    1 0 0 | 0.8
    1 0 1 | 0.8
    1 1 0 | 0.8
    1 1 1 | ? (maybe you can figure out the result here?)

    Y, is the number of open paths (0, 1 or 2 paths 'open', with 0.8 probability of actually opening)
    Whoops, that should have said: "with each valve in any path having a 0.8 probability of actually opening".
     
    Last edited: Oct 2, 2008
  8. Pete It's not rocket surgery Registered Senior Member

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    10,167
    The "signal" is just whatever is supposed to open the valves, e.g. a button labeled "open valves 1,2,3".

    The valves are unreliable - when the button is pushed, each valve has a 20% chance of sticking shut (ie the probability that a valve opens when the button is pressed is 0.8).

    So... when you push the button, what is the probability that both the top and bottom pipes will be fully open?
    What is the probability that only one of the paths will be open?
    What is the probability that both paths will be blocked?
     
  9. kingwinner Registered Senior Member

    Messages:
    796
    I think the probability that both the top and bottom pipes will be fully open = 0.8^3
    But I have some trouble with your two latter questions, how to find the probability?
     
  10. kingwinner Registered Senior Member

    Messages:
    796
    I am a bit confused...what does B represent? Is it the probability? If so, wouldn't it be:

    signal | B (output)
    0 0 0 | 0.2^3 ?
    0 0 1 | 0.8 x 0.2^2 ?
    0 1 0 | 0.8 x 0.2^2 ?
    0 1 1 | 0.8^2 x0.2 ?
    1 0 0 | 0.8 x0.2^2 ?
    1 0 1 | 0.8^2 x 0.2 ?
    1 1 0 | 0.8^2 x 0.2 ?
    1 1 1 | 0.8^3 ?
     
  11. Vkothii Banned Banned

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    3,674
    Well, the diagram in post #1, has an input A, and an output B. Notice I labelled B with the word "output"?

    Can you figure this question out about the circuit?
    When all valves are closed, how many paths are there open from A to B?
    My guess is "none"
    Why have you cubed B? What's the idea there?

    (Edit)
    Perhaps it might help to decompose the problem (an approach that's fairly standard), simplify things to start with.

    Assume there is one path, and one valve. If the valve is opened (with an 80% probability), how many paths are open?
    One path, one valve, one probability (0.8). The output (B) is therefore 0.8 for a single valve, you agree?

    Now try two valves; for there to be a path both have to open, right? So the probability is 0.8 x 0.8 for two valves that have a 0.8 probability of opening, agree?

    If you can get from one path with one valve, to one path with two valves, perhaps you can get to one path with one valve, plus one path with two valves (i.e two paths)?
    How about it?
     
    Last edited: Oct 4, 2008
  12. Pete It's not rocket surgery Registered Senior Member

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    10,167
    Vkothkii, your table doesn't make much sense.
    The output is either zero paths open, one path open, or two paths open. An output of 0.8 isn't an option.

    You're on the right track with your change to Vkotthi's table to show the probability of each configuration. Add another column to that table to show how many paths are open for each configuration, then you're pretty much done!
    Code:
     Valves   |               |       |
     1  2  3  |  Probability  | Paths |
    ----------+---------------+-------|
     0  0  0  |    0.2^3      |       |
     0  0  1  |  0.8 x 0.2^2  |       |
     0  1  0  |  0.8 x 0.2^2  |       |
     0  1  1  |  0.8^2 x 0.2  |       |
     1  0  0  |  0.8 x 0.2^2  |       |
     1  0  1  |  0.8^2 x 0.2  |       |
     1  1  0  |  0.8^2 x 0.2  |       |
     1  1  1  |     0.8^3     |       |
    ----------------------------------
    
     
  13. Vkothii Banned Banned

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    3,674
    I agree, with two paths there can only be a maximum of two open, and a minimum of zero open. One is the only available number of open paths between zero and two. That's logic for ya.
    If there is a single path, with a single valve in it, what's the probability of there being an output, if the valve opens with a probability of 0.8? Is there an assumption that a closed valve has a 20% probability of being open? Where does it say that in the first post?

    Can you explain to a really dumb person like me, why there is a probability of 0.2 x 0.2 x 0.2 of an output, with no valves open?

    You think my table doesn't make much sense, whereas I don't think the table in #7 or #9 makes much sense. Can you explain what the logic is, how you got to that list of probabilities?
     
    Last edited: Oct 4, 2008
  14. Vkothii Banned Banned

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    3,674
    Never mind, here's my solution:

    with all valves shut (assuming they operate correctly, on signal, and that a '0' signal means the valve is fully closed, a '1' signal means it opens with a probabilty of 0.8, or a probabilty of 0.2 of staying closed).

    Code:
    valve#| output| #paths open
    1 2 3 
    0 0 0 | 0.00 | 0
    0 0 1 | 0.00 | 0
    0 1 0 | 0.00 | 0
    0 1 1 | 0.64 | 1
    1 0 0 | 0.80 | 1
    1 0 1 | 0.80 | 1
    1 1 0 | 0.80 | 1
    1 1 1 | 0.72 | 2 (..oops)
     
    Last edited: Oct 4, 2008
  15. Vkothii Banned Banned

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    3,674
    "Find the probability distribution / probability mass function for \(Y\), the number of open paths from A to B after the signal is given; \(Y \in {0, 1, 2}.\)"

    Ok.
    This problem illustrates encoding with a matrix; it's a gating or computational exercise: - A is 'cloned' or copied (a fan-out), into two parallel inputs.

    There are two different 'channels' which encode the information (water, in this case) statistically; the channels produce 'mixed' states for the two inputs, which get 'measured/sampled' at B, the joint output, an eigenvalue for the circuit function.

    P.S. in theory, it models a photon going through a half-mirror, or a down-converter, then the entangled pairs traveling along different paths, and recombining at a detector - there are three allowed phases for the output, in terms of the settings or states of each valve - the valves could be polarising filters.

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    Last edited: Oct 4, 2008
  16. Pete It's not rocket surgery Registered Senior Member

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    There is a probability of 0.2x0.2x0.2 that all three valves will stay closed. If that happens, no paths will be open - the output is zero.

    There are no "zero" signals and "one" signals... there is a single signal.
    The result of that signal should be that all the valves open, but for each valve there is a 0.2 chance that it will stay closed.
     
  17. kingwinner Registered Senior Member

    Messages:
    796
    Thanks for helping!

    Code:
     Valves   |               |       |
     1  2  3  |  Probability  | Paths |
    ----------+---------------+-------|
     0  0  0  |    0.2^3      |    0   |
     0  0  1  |  0.8 x 0.2^2  |    0   |
     0  1  0  |  0.8 x 0.2^2  |    0   |
     0  1  1  |  0.8^2 x 0.2  |    1   |
     1  0  0  |  0.8 x 0.2^2  |    1   |
     1  0  1  |  0.8^2 x 0.2  |    1   |
     1  1  0  |  0.8^2 x 0.2  |    1   |
     1  1  1  |     0.8^3     |    2   |
    ----------------------------------
    
    P(Y=0)=0.2^3 + 0.8 x 0.2^2 + 0.8 x 0.2^2
    P(Y=1)=0.8^2 x 0.2 + 0.8 x 0.2^2 + 0.8^2 x 0.2 + 0.8^2 x 0.2
    P(Y=2)=0.8^3

    Is this correct?
     
  18. Pete It's not rocket surgery Registered Senior Member

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    10,167
    It's the same as what I got, at least.
    You should do a quick reality check and make sure that the three probabilities add up to 1.
     
  19. Vkothii Banned Banned

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    3,674
    That's what the OP says - the valves are independent; there are 3 valves that can be open or closed. How many combinations is that?
    Or there are 3 independent signals for the independent valves?

    If the 'signal code' is 111, all 3 valves will open independently 80% of the time. Or stay closed 1 in 5 times they all get told to open together, or as you yourself say "for each valve there is a 0.2 chance that it will stay closed"; being independent, there's an independent code, OK? A time-dependent one, like most other codes, but that's what the problem is about. Encoding a signal. It also underlines nicely how information is physical.

    Or show how it doesn't mix two parts of a signal, how there are not 3 states, like I said, measured at B, in terms of say "the probability of the water input at A"?

    P.S. Doing it your way, with a single "on" signal, means each of eight possible combinations of valve settings - 'open/closed', is the equivalent of a time dependent code. Each combination has an equal probability, rather than what your table shows(?) How does each state for the independently operating valves have a different probability. if it means each time the "on" button is pushed, there are eight possible outcomes for the 3 valves?
     
    Last edited: Oct 4, 2008
  20. Pete It's not rocket surgery Registered Senior Member

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    10,167
    No, I think you're making it too complicated, and reading things into the question that aren't there. If it were a complex signal rather than a simple "on", then I really think there would be more description of the signal in the question.

    The problem is about independent probabilities. It's high school or first year college maths.


    But, I concede that there is room for interpretation. The question could have been stated more clearly.
     
  21. Vkothii Banned Banned

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    3,674
    I think you're missing the point of the question. What happens if each of 3 valves, as configured, opens 50% of the time they get their signal (to open)?
     
  22. Pete It's not rocket surgery Registered Senior Member

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    10,167
    Then it's the same problem, except that you change all the 0.8's and 0.2's to 0.5's when calculating the answer.
     
  23. Vkothii Banned Banned

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    3,674
    Assuming the valves operate correctly, as the OP states, means you assume they open 80% of the time, then with a '111' code there are eight possibilities, a '000' code means 1 possibility = all valves stay closed.
    Assume a '100' code means open valve 1. There are 2 possible outcomes with a 0.8 'average over time/number of signals received' of it opening. The possibility of water at B in terms of A is then 80% (over time).

    The connection is between which signal codes result in which settings for the valves, and what the resulting open number of paths is. There are operations on the 'code sets' to get from one Y to another.
    It is ambiguous because you can assume a 3-bit code, and a 3-bit 'probable' outcome for each code.

    But it's about codes, encoding and transforms; and it's analogous to signalling - water is information, information is physical - fan-out or copy increases information content or entropy.
    It encapsulates a few things.
     
    Last edited: Oct 4, 2008

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