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B Probability question

  1. Jun 27, 2016 #1
    How do I express the following scenario mathematically?

    I have access to CNN news online from two different internet sites, each of which have about an 80% reliability of actually providing the feed when I log onto the site. If I only had access to one of the sites, I'd know that I had an 80% chance that I'd get a live feed when I logged in. How does this figure change when I have two sites, both with that 80% reliability.

    The temptation is to multiply the two probabilities together, i.e., .8 * .8 = .64 But that can't be right because the probability has to be greater than .8

    Can someone set this scenario up for me mathematically?
  2. jcsd
  3. Jun 27, 2016 #2


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    If you multiply them, you're asking for the probability of getting a news feed from one site and then the other.
    I think I'm missing some detail, though. So you have to choose a site, and then log in? Or what? Do both sites pop up when you log onto whatever you're logging into?
  4. Jun 27, 2016 #3


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    If that's the case, you're looking for the probability of A and/or B = A or B + A and B. At least I would think so. You only care that you get at least one, but it's also a possibility that you get two, so you need the probability of getting 1 + the probability of getting 2.
  5. Jun 27, 2016 #4


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    Ok, so this is what I'm thinking, I get a value of 96% when I do this.
    ##A## is getting the feed from A ##A'## is not getting the feed from A.

    ##P = AB' + A'B + AB##
  6. Jun 27, 2016 #5


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    Let [itex]A[/itex] be the outcome "The first site works". Let [itex]B[/itex] be the outcome "The second site works". Then there are 4 possible joint outcomes: [itex]A \wedge B, A \wedge \neg B, \neg A \wedge B, \neg A \wedge \neg B[/itex] (where [itex]\neg[/itex] means "not", and [itex]\wedge[/itex] means "and"). So the probability that at least one of the sites works is:

    [itex]P(A \vee B) = P(A \wedge B) + P(A \wedge \neg B) + P(\neg A \wedge B)[/itex]

    where [itex]\vee[/itex] means "or".

    So do you know how to compute those three probabilities on the right side of the equals?
  7. Jun 27, 2016 #6
    No I don't. But for some reason I thought intuitively the answer was what Bigyellowhat stated, 96%

    I don't know where I came up with that figure, and when I tried to think about how the probability would be characterized mathematically, I drew a blank. This is especially disturbing since I just took the GRE last Summer and had these type of probability questions drilled into my head. I've already forgotten how to do them :oldconfused:
  8. Jun 27, 2016 #7


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    What's the probability of A but not B?

    As for where I got that number, it's the same as what Steven Darryl is doing.
  9. Jun 27, 2016 #8


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    You have to use some laws of probability:
    [itex]P(\neg X) = 1 - P(X)[/itex]
    [itex]P(X \wedge Y) = P(X) \cdot P(Y)[/itex] (if the probabilities are independent)
  10. Jun 27, 2016 #9
    That looks like it adds up to .96! If there's one thing I still know how to do, it's add decimals o0)

    .64 + .16 + .16 =.96 Am I right? I think the probabilities are independent. If you have cable or satellite, you pretty much have a 100% reliability that you will get a CNN feed. But the two sites I get my CNN news feed from have essentially an 80% reliability independent of one another. I was just wondering how much closer to 100% my reliability was by having the two options.
  11. Jun 27, 2016 #10
    These answers seem somewhat over-complicated. Assuming the probabilities of NOT getting a feed from one site are independently 1 - 0.8 = 0.2, the probability of not getting a feed from both sites is 0.2 x 0.2 = 0.04. So the probability of this not happening (i.e. getting a feed from at least one site) is 1 - 0.04 = 0.96.

    [edit - added this]And if you have three sites available, the probability of not getting a feed from any is 1 - 0.23 = 99.2%.
  12. Jun 27, 2016 #11


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    Yes, it's 96% if the two are independent. The simplest calculation is how often you have neither. That's 0.2 x 0.2 = 0.04.

    0.8 x 0.8 = 0.64 is the probability you have both.

    And 0.8 x 0.2 = 0.16 is the probability you have the first but not the second. Similarly it's the same probability you have the second but not the first. Hence:

    64% both feeds available
    32% only one feed available
    4% neither feed available
  13. Jun 27, 2016 #12


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    I suppose I overlooked that, no? lol
  14. Jun 27, 2016 #13
    I like how you broke that down PeroK, thanks.
  15. Jun 27, 2016 #14
    This is studied in reliability analysis. Basically, you have some subdevices (here: the two sites) which are operational which probability ##p##. What is asked here is what is the probability that the entire device is operational.

    If the devices are connected in series, then the probability is ##p^n## where ##n## is the number of devices.
    If the devices are connected in parallel, then the probability is ##1 - (1- p)^n##.
    You can generalize this to ##k## out of ##n## systems where the device works if ##k## out of ##n## subdevices work. This requires the binomial distribution. Of course, you can go even more complicated than this.
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