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Confusion with probability models and intersection of events

  1. Sep 22, 2012 #1
    1. The problem statement, all variables and given/known data

    I will use an example to showcase my confusion:

    Suppose a person watches show A 2/3 of the time, show B 1/2 of the time, and both show A and show B 1/3 of the time. For a randomly selected day, what is the probability that the person watches only show A? For a randomly selected day, what is the probability that the person watches neither show?

    2. Relevant equations

    A probability measure [itex]P[/itex] is such that if [itex]A_1, A_2, \cdots[/itex] is a finite or countable sequence of pairwise disjoint events, then [itex]P(A_1 \cup A_2 \cup \cdots) = P(A_1) + P(A_2) + \cdots[/itex]

    3. The attempt at a solution

    Maybe I'm looking too deeply, or I'm just confused, but here it is...

    My first concern is that, if I define [itex]A[/itex] to be the event "the person watches show A," and [itex]B[/itex] the event "the person watches show B," then I have [itex]P(A) + P(B) > 1[/itex]. (These numbers are what was provided.) Does this mean that they cannot be part of the same sample space? Is that okay?

    My second concern is that, to my understanding, the required event is [itex]A \cap B^C[/itex], where [itex]B^C[/itex] is the complement of [itex]B[/itex]. To find this, we can write [itex]A = (A \cap B^C) \cup (A \cap B)[/itex] and then apply the additivity property due to how they are disjoint. But, contrary to how the problem has [itex]P(A \cap B) = 1/3[/itex], in terms of set theory, wouldn't this be [STRIKE]∅[/STRIKE] [itex]P(∅)=0[/itex], due to how [itex]A[/itex] and [itex]B[/itex] are disjoint? In my textbook, there was never a formal definition of what the intersection of "events" is, but in my class our professor defined it, maybe informally, to be "the event that both events occur." What definition should I be using?

    Anyway, if I go along and solve it ignoring these aspects, you get 1/3 as the probability measure of the first required event, i.e. of [itex]A \cap B^C[/itex].

    Moving onto the second question, I become confused. Would this event (the event the person watches neither shows) be [itex]A^C \cap B^C[/itex], leaving room for other possible events not discussed, or just ∅, if I make the sample space only consisting of these events, even though that brings me back to my first problem?

    Thanks in advance.
     
  2. jcsd
  3. Sep 22, 2012 #2

    jbunniii

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    It's OK that the probabilities add up to more than one. This isn't a contradiction of anything. Note that events A and B are not disjoint, i.e., they overlap. So when you add P(A) + P(B), you are counting the overlap (which is [itex]P(A \cap B)[/itex]) twice. In general, you have
    [tex]P(A \cup B) = P(A) + P(B) - P(A \cap B)[/tex]
    which removes the double-counting of the overlap.
    Correct.
    What makes you say that [itex]A[/itex] and [itex]B[/itex] are disjoint? They aren't. If they were disjoint, then you would have [itex]P(A \cap B) = 0[/itex].
    Correct.
    Yes. Notice that this is the complement of [itex]A \cup B[/itex].
    I really don't understand what you're asking here. Can you maybe give an example of what you mean?
     
  4. Sep 22, 2012 #3

    HallsofIvy

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    No, it means that they are NOT "pairwise disjoint" or, in probability terms, they are not
    "mutually exclusive". It is possible that the person watches both A and B. In fact, you are told that " both show A and show B 1/3 of the time." [itex]P(A\cup B)= P(A)+ P(B)- P(A\cap B)[/itex] which here is 1/2+ 2/3- 1/3= 1/2+ 1/3= 5/6.[/itex]


    My second concern is that, to my understanding, the required event is [itex]A \cap B^C[/itex], where [itex]B^C[/itex] is the complement of [itex]B[/itex]. To find this, we can write [itex]A = (A \cap B^C) \cup (A \cap B)[/itex] and then apply the additivity property due to how they are disjoint. But, contrary to how the problem has [itex]P(A \cap B) = 1/3[/itex] , in terms of set theory, wouldn't this be [STRIKE]∅[/STRIKE] [itex]P(∅)=0[/itex], due to how [itex]A[/itex] and [itex]B[/itex] are disjoint?[/quote]
    A and B are NOT disjoint- again, you are told that- [itex]P(A\cap B)= 1/3[/itex].

    The "set of outcomes" is the set of all individual things that could happen- the "possible outcomes". An "event" is a subset of that- a set containing some of the things that could happen. The intersection of two events is the intersection of those two sets.

    No. "not watching either A or B" is also an event, having non-zero probability.

    You are given that the probability that the person watches program A is P(A)= 2/3, that the probability that the person watches program B is P(B)= 1/2, and that the probability that the person watches both is [itex]P(A\cap B)= 1/3[/itex]. So the probability that the person watchs "A or B or both" is [itex]P(A\cup B)= P(A)+ P(B)- P(A\cap B)= 2/3+ 1/2- 1/3= 1/3+ 1/2= 5/6[/itex] as before. The probability that the person did not watch either A or B is 1 minus that.
     
  5. Sep 22, 2012 #4

    Ray Vickson

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    Try drawing Venn diagram. Until you get comfortable with this type of material, such a diagram helps to direct your thinking along appropriate lines.

    RGV
     
  6. Sep 22, 2012 #5
    I don't understand why they are not disjoint... I think that's my main problem.

    The following is how I see it. If I flip a fair coin, then there are two events [itex]\left\{H\right\}[/itex] and [itex]\left\{T\right\}[/itex]--oh, I think I see what's happening. Here, the coin can't be both heads and tails, because it follows from it being assumed to be fair. However, this isn't a general thing I should assume. Specifically, I shouldn't assume that the person in the OP example can't both watch show A and show B, as this is intuitively implied. Do I have the right idea?
     
  7. Sep 22, 2012 #6

    jbunniii

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    They are not disjoint because it's possible for both of them to happen. From the original problem statement: "...and both show A and show B 1/3 of the time." If two events are disjoint, then they cannot both happen. This is the definition of disjoint.

    Expressing it in terms of probability, if A and B are disjoint, then [itex]P(A \cap B) = 0[/itex]. In this problem, you have [itex]P(A \cap B) = 1/3[/itex], so A and B are not disjoint.

    Yes, that's correct. A single coin toss cannot result in both H and T. Whereas it's possible to watch both shows A and B in the same day, and in fact this happens with nonzero probability in this problem.
     
  8. Sep 22, 2012 #7

    Ray Vickson

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    In the coin case it has nothing at all to do with the fairness or not of the coin; it just has to do with the fact that when you toss a coin you cannot have BOTH heads and tails at the same time. In the TV shows example you were told explicitly that both A and B can happen together, and you were even told the probability of that happening.

    You would not have such trouble understanding the problem if you just read it carefully and think about what it is saying.

    RGV
     
  9. Sep 24, 2012 #8
    Okay... I think I've got a better understanding, now.

    Thank you for your help (and every else).
     
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