Probability for a rule of 7

Discussion in 'PMP' started by Sahil Mahajan_1, Jun 13, 2019.

  1. Sahil Mahajan_1

    Sahil Mahajan_1 Well-Known Member

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    Could someone please provide the correct answer to below question, along with explanation?

    By the rule of seven, a process is said to be out of control if a run of seven samples is found on one side of the process mean. What is the probability that a run of seven occurs on either side of the mean due to random variation?
    SELECT THE CORRECT ANSWER


    A. 0.02
    B. 0.035
    C. 0.0156
    D. 0.0273
     
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  2. tim jerome

    tim jerome Well-Known Member
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    If you flip a coin 7 times, the odds of getting 7 heads in a row is 1 in 2^7 or 1 in 128.

    This is if everything is 'fair and equal', or "random". This works out to be about .78% probability.

    You can arrive at this from a more step-wise manner -

    Each coin flip is in isolation. Each coin flip is 50% Heads before you flip it. Therefore,
    the chance two heads in a row occur = 50% * 50%, or (.5)^2 = 25%. Every time you add another
    coin flip, this is modified by another 50% probability - the third flip multiplies this result,
    and the 4th modifies that one. Therefore, the answer can be reached by calculating (.5)^7.

    Does anyone else see this differently?

    I'd like to see the answer and its description.
     
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  3. Sahil Mahajan_1

    Sahil Mahajan_1 Well-Known Member

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    Thanks Tim,
    I also thought of the same logic. However the answer mentioned in Simplilearn Simulation exam is (0.5)^6. I do not understand the explanation provided there though. It was that the first point will have a probability of A and next six points each would have a probability of (1/2A). This does not make sense to me.
     
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  4. tim jerome

    tim jerome Well-Known Member
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    I found another comment online, which is interesting (I continue to research this problem).

    The probability that we have a run of 7 above the mean is (.5)^7.

    The probability that we have a run of 7 below the mean is (.5) ^7

    The probability that we have a run of 7 either below or above the mean is (.5)^6.

    This makes sense - the third option considers the first coin flip to establish which side the run of 7 is to occur, the rest evaluate the probability.

    I think I agree with this, but I need to work through the logic a few more times.

    Does this help?
     
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  5. Sahil Mahajan_1

    Sahil Mahajan_1 Well-Known Member

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    Sorry Tim, not very clear. Would you be able to explain this more?
     
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  6. tim jerome

    tim jerome Well-Known Member
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    So - every time you flip a coin, you add another chance of adding to the string of heads.

    This is why a string of 7 heads in a row can be thought of as (50%)^7 - a random chance of being either heads or tails.

    The scenario states, however, that we are calculating a run of 7 either above or below our target, or mean. This in effect says, "let's disregard the probability of the first coin flip, and merely use it as the point where the run commences."

    Therefore, a run of 7 has one coin flip that establishes the run, where the rest add a 50% probability to the total probability of the run - (50%)^(7-1), or (50%)^6.
     
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  7. Sahil Mahajan_1

    Sahil Mahajan_1 Well-Known Member

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    Got you. Thank you for the explanation Tim!
     
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