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Homework Help: Probability and De Morgan's laws question.

  1. Mar 6, 2013 #1
    Hi,
    1. The problem statement, all variables and given/known data
    A coin is tossed three times. I am asked for the probability that EITHER the first toss yields heads, OR the second toss yields tails, OR the third yields heads. I am expected to use De Morgan's Laws but am not sure how to define the events the themselves. I'd appreciate some guidance.


    2. Relevant equations



    3. The attempt at a solution
    I have found the reuested probability using the following table
    First toss: HTT
    Second toss: HTH
    Third toss: TTH
    to be 3/8.
    But how may I show that rigorously? Mind you, we haven't dealt with multiplication of probabilities (so (1/2)3 for each case, let's say, would not be legit). We have hitherto only dealt with De Morgan's Laws, P(AUB)=P(A)+P(B)-P(A and B), and P(AUBUC)=P(A)+P(B)+P(C)-P(A and B) - P(B and C) - P(A and C) + P(A and B and C).
    As mentioned, I'd appreciate your assistance.
     
  2. jcsd
  3. Mar 6, 2013 #2

    haruspex

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    Suppose A, B, C are the events that, respectively, 1st, 2nd 3rd tosses produce a H. Write A' etc. for tail event. Can you write the expression for the desired compound event?
     
  4. Mar 6, 2013 #3
    Hi haruspex,
    Following your notation, I believe the probability should hence be P(A U B' U C) = P(A) + P(B') + P(C) - P(A and B') - P(B' and C) - P(A and C) + P(A and B' and C), but this yields 7/8!
     
  5. Mar 6, 2013 #4
    Whereas the correct answer should be 3/8, should it not?
     
  6. Mar 6, 2013 #5

    Ray Vickson

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    If you have not yet had 'multiplication of probabilities', what are you expected to do? Somehow, you need to use the fact that the outcome of the first toss does not affect the chances of the outcomes on the second toss, and so forth. You *could* argue that the probability of the event you are given is the same as the probability of H on toss 1 OR H on toss 2 OR H on toss 3. It might be easier to see what is the probability of this latter event.
     
    Last edited: Mar 6, 2013
  7. Mar 6, 2013 #6
    First of all, could someone please clarify whether the answer should be 3/8 or 7/8?
     
  8. Mar 6, 2013 #7

    Ray Vickson

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    First you need to explain how you got your answer, since you claim you cannot use the "multiplication of probabilities".

    Anyway, I don't think we are allowed to tell you exactly what the answer is; we are only allowed to supply some hints.
     
  9. Mar 6, 2013 #8
    Well, 3/8 was derived using multiplication, and as delineated above in my first post to this thread, which is why I decided to ask for assistance here as I am unable to arrive at that answer without resorting to multiplication and using De Morgan's Laws.
    7/8 was arrived at using the formula I noted above, and presumably following haruspex's rationale, again using multiplication.
    Would you be more inclined to assist me now?
     
  10. Mar 6, 2013 #9

    haruspex

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    Two hints that might help you decide that:
    - would HHH be a 'success'?
    - what combinations of toss outcomes would not be a success?
     
  11. Mar 6, 2013 #10

    Ray Vickson

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    I tried to: I said:
    "You *could* argue that the probability of the event you are given is the same as the probability of H on toss 1 OR H on toss 2 OR H on toss 3. It might be easier to see what is the probability of this latter event."
     
  12. Mar 6, 2013 #11
    There is more than one combination which would not be a success, so probability (I think) should be 3/8. For instance, both HHH and TTH would not be a success. Am I right?
     
  13. Mar 6, 2013 #12
    The only successful combinations are:
    HHT, TTT, and THH.
    Unless I am mistaken.
     
  14. Mar 6, 2013 #13

    haruspex

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    You are mistaken. Success was defined as "EITHER the first toss yields heads, OR ... OR ..." . A sequence of HHH satisfies "the first toss yields heads". No need to check any further.
     
  15. Mar 6, 2013 #14
    But it is "OR"! Hence, if you first get heads then the third cannot also be heads. I believe I am right. Question is how to demonstrate that using De Morgan's Laws.
     
  16. Mar 6, 2013 #15

    Ray Vickson

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    NO, no, no! OR mean either/or, so *both* can happen. Please jetisson right now any ideas you may have that OR means 'one or the other but not both'; if we want that, we say it explicitly, or else say something like "XOR".

    Don't take my word for it; look it up on-line or in a textbook.
     
  17. Mar 6, 2013 #16
    Even if the word "OR" in the question itself was put in bold, so as to, apparently, imply "XOR"?
     
  18. Mar 6, 2013 #17

    Curious3141

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    When we use "OR" in the English language, the most common implication is as "XOR", e.g.

    "You can have coffee OR tea" generally excludes the option of having both beverages.

    When you want to imply the Boolean logic understanding of "OR", the usual convention is to state "OR both": e.g. "You can have coffee OR tea OR both".

    Mathematical word problems (like this one) are written in English, so you still need to interpret them accordingly.

    (EDIT: However, the correct interpretation of this (as with most other probability problems) involves a Boolean OR)

    Draw a tree diagram to represent events here (see attachment).

    The circles in red are the only events that you're interested in. Remember to apply the laws of conditional probability as you go through the tree (this involves multiplication because of the implicit "AND" condition).

    That diagram pretty much represents the exact understanding a lay speaker of English would have of events. EITHER the first toss is heads OR ... (and at this point, you consider only events stemming from the first toss coming up tails), and so forth.
     

    Attached Files:

    Last edited: Mar 7, 2013
  19. Mar 6, 2013 #18

    Ray Vickson

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    Bold would not do it, but maybe---just maybe---the word "either" might. However, I have seen hundreds of questions like this in which one is asked for the probability of either this or that or something else, and "ordinary" OR was meant. That is standard language.

    However, not being clairvoyant, I cannot say with certainty what the questioner meant.
     
  20. Mar 7, 2013 #19

    Curious3141

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    As I wrote, it doesn't actually matter for this problem.

    EDIT: It actually matters, and this is using a Boolean "OR" as is the usual convention.
     
    Last edited: Mar 7, 2013
  21. Mar 7, 2013 #20

    Ray Vickson

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    Actually, it matters a lot, because in PROBABILITY (rather than standard English) the usual interpretation of "OR" is the inclusive one---either one or the other or both.

    Of course, some types of dichotomies are automatically XOR, such as getting either a head or tail in a single toss, or a subject being either male of female, or whatever. However, when it comes to "compound" events that is often not the case. Let me repeat: I have seen hundreds of questions of this type in which the word "OR" means the inclusive one---no lie. In some other areas of discourse, maybe things are different, but not usually in probability.

    However, perhaps the OP knows what his instructor or textbook usually means by OR, and he can use that.
     
  22. Mar 7, 2013 #21

    Curious3141

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    (never mind - please see my latest post)
     
    Last edited: Mar 7, 2013
  23. Mar 7, 2013 #22

    haruspex

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    The everyday meaning of OR varies according to context. If I ask you whether you have change of $5 or $10, you won't answer no just because you have both. If you place a bet with a bookmaker that a team will win either of its next two matches, you would be upset to be refused a payout when it wins both.
    Where 'both' is excluded, it's usually (always?) pretty obvious from the circumstance - such as, offering tea or coffee. If I ask whether you drink tea or coffee, suddenly it's different.
    In the OP context, it seems to me quite unreasonable to interpret it as exclusive.
     
  24. Mar 7, 2013 #23

    Curious3141

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    (never mind - please see my latest post)
     
    Last edited: Mar 7, 2013
  25. Mar 7, 2013 #24

    Curious3141

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    My apologies - ignore what I'd written with regard to the OR/XOR ambiguity. I was not thinking straight previously with a bad head cold.

    The tree diagram is correct, and I realise that this is using a simple Boolean OR, which is how I read the question now that I feel a little more clear-headed.

    Reading it as an XOR would exclude an event like Toss 1 = Head AND Toss 2 = Head AND Toss 3 = Head (for example), whereas the question clearly intends for this case to be counted as an event based on the first line.

    So, peripatein: this OR/XOR differentiation is important, and here (as usual), the Boolean OR interpretation should be employed.

    I read the question correctly throughout, but misinterpreted the logical relationship - it should be an "OR" as Ray and Haruspex correctly pointed out.
     
    Last edited: Mar 7, 2013
  26. Mar 7, 2013 #25

    Ray Vickson

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    Let me first emphasize that I think the following will be answering the wrong question, but here goes anyway.

    For events A = first toss is H, B = second toss is T, C = third toss is H, how do we get P{A xor B xor C} from standard probability and set manipulation laws? We have
    [tex] A \text{ xor } B \text{ xor } C = [A \cap (B \cup C)^c ] \cup [B \cap (A \cup C)^c]
    \cup [C \cap (A \cup B)^c], [/tex] and the three events in square brackets are mutually exclusive, so P(A xor B xor C) is the sum of their probabilities. We have
    [tex] A = [A \cap (B \cup C)] \cup [A \cap (B \cup C)^c],[/tex] and the two events on the right are mutually exclusive, so
    [tex] P(A) = P(A \cap (B \cup C)) + P(A \cap (B \cup c)^c), \text{ or}\\
    P(A \cap (B \cup C)^c) = P(A) - P(A \cap (B \cup C)) = P(A) - P[(A\cap B) \cup (A \cap C)] \\
    = P(A) - P(A \cap B) - P(A \cap C) + P[(A \cap B) \cap (A \cap C)] \\
    = P(A) - P(A \cap B) - P(A \cap C) + P(A \cap B \cap C)
    = 1/2 - 1/4 - 1/4 + 1/8 = 1/8.[/tex]

    Of course, you can also see this directly, since "A but not B and not C" means the first toss is H, the second is not T (so is H) and the third toss is not H (so is T); that is, it is the single outcome HHT.
     
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