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    (Original post by AFraggers)
    Yeah I kind of understand.If you were doing this question using the inclusion- exclusion formula, how would you get to your answer??
    If you've never come across it, don't worry. I've only come across it this semester as a second year Maths undergrad.

    But you use the formula, so P(AUBUC) = P(A) + P(B) + P(C) - P(ANB) - P(ANC) - P(BNC) + P(ANBNC).

    We know that AUBUC = S so P(AUBUC) = 1. We know what the other things are except P(C), so take everything away from both sides leaving P(C) on the right and on the left P(AUBUC) - P(A) .... - P(ANBNC), and P(ANBNC) = 0 since P(empty) = 0.

    That is basically what you should do with the whole shading / double counting thing but uses a formula.
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    (Original post by SeanFM)
    If you've never come across it, don't worry. I've only come across it this semester as a second year Maths undergrad.

    But you use the formula, so P(AUBUC) = P(A) + P(B) + P(C) - P(ANB) - P(ANC) - P(BNC) + P(ANBNC).

    We know that AUBUC = S so P(AUBUC) = 1. We know what the other things are except P(C), so take everything away from both sides leaving P(C) on the right and on the left P(AUBUC) - P(A) .... - P(ANBNC), and P(ANBNC) = 0 since P(empty) = 0.

    That is basically what you should do with the whole shading / double counting thing but uses a formula.
    Using the values from the original question, Do you get an answer of 0.897???
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    (Original post by AFraggers)
    Using the values from the original question, Do you get an answer of 0.897???
    For P(C)? That seems a bit too high.

    Actually , with the values and that formula that does look like something you'd get.
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    (Original post by SeanFM)
    For P(C)? That seems a bit too high.

    Actually , with the values and that formula that does look like something you'd get.
    ok i'll assume it's right. Thanks for the help!
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    (Original post by AFraggers)
    ok i'll assume it's right. Thanks for the help!
    I get the feeling you're not supposed to know that formula.

    As long as you understand why it's the same as looking at the Venn diagram, and what makes up P(A) for example.

    In a more simple example, where A and B are two overlapping circles on a Venn diagram, if you had P(AUB) = 1, P(A) = 0.7, P(ANB) = 0.2 then on the Venn diagram you would have in A (not overlapping with B) 0.7-0.2 (0.5), 0.2 in the middle and to find P(B but not A) it would be 1 - 0.5 - 0.2 = 0.3, because you've already accounted for the overlapping, but this is probably confusing.
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    (Original post by SeanFM)
    I get the feeling you're not supposed to know that formula.

    As long as you understand why it's the same as looking at the Venn diagram, and what makes up P(A) for example.

    In a more simple example, where A and B are two overlapping circles on a Venn diagram, if you had P(AUB) = 1, P(A) = 0.7, P(ANB) = 0.2 then on the Venn diagram you would have in A (not overlapping with B) 0.7-0.2 (0.5), 0.2 in the middle and to find P(B but not A) it would be 1 - 0.5 - 0.2 = 0.3, because you've already accounted for the overlapping, but this is probably confusing.
    In terms of an alternative method, how would you answer this question without the formula???
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    (Original post by uAFraggers)
    In terms of an alternative method, how would you answer this question without the formula???
    Basically the same method but without the formula, which is what I was trying to explain/hint at in the previous post.

    Once you've got the correct Venn diagram all you need to do is 1 - all of the numbers to find P(C on its own) and then add on the other areas involving C to find P(C) (As P(C) = P(C on its own) + P(A intersect C) + P(B intersect C) + P(A intersect B intersect C).

    That formula is how you get the 'correct' value of P(A on its own) as you are given everything else in the question
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    (Original post by SeanFM)
    ..
    Do you mind helping me with a question.

    7i. I have done pretty much all of it. Though I am unsure on when you use the central limit theorem thing.

    Like I thought since it was a sample you have to divide the standard deviation by n.

    thanks
    http://www.ocr.org.uk/Images/242460-...atistics-2.pdf
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    (Original post by Super199)
    Do you mind helping me with a question.

    7i. I have done pretty much all of it. Though I am unsure on when you use the central limit theorem thing.

    Like I thought since it was a sample you have to divide the standard deviation by n.

    thanks
    http://www.ocr.org.uk/Images/242460-...atistics-2.pdf
    CLT is used when you're using the normal distribution.

    In this case I think you're fine without it as you are directly doing it with binomial

    If something is distributed normally with mean mu and standard deviation sigma, then the mean of n of those things is distributed normally with mean mu and standard deviation sigma/(sqrt(n)).
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    (Original post by SeanFM)
    CLT is used when you're using the normal distribution.

    In this case I think you're fine without it as you are directly doing it with binomial

    If something is distributed normally with mean mu and standard deviation sigma, then the mean of n of those things is distributed normally with mean mu and standard deviation sigma/(sqrt(n)).
    Am I?
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    (Original post by Super199)
    Am I?
    I could've sworn you wrote 6i

    Let me look again
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    (Original post by SeanFM)
    I could've sworn you wrote 6i

    Let me look again
    Yeah I did accidentally haha my bad.
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    (Original post by Super199)
    Do you mind helping me with a question.

    7i. I have done pretty much all of it. Though I am unsure on when you use the central limit theorem thing.

    Like I thought since it was a sample you have to divide the standard deviation by n.

    thanks
    http://www.ocr.org.uk/Images/242460-...atistics-2.pdf
    The CLT is used when you're using the distribution of the mean. If you're using it then you don't need to assume that the data is normally distributed (7ii) because the CLT says that it is, provided it is a large sample.

    But in i) though you can't use the CLT, by using the normal distribution to estimate the proportion you are assuming that the data is normally distributed.
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    (Original post by SeanFM)
    The CLT is used when you're using the distribution of the mean. If you're using it then you don't need to assume that the data is normally distributed (7ii) because the CLT says that it is, provided it is a large sample.

    But in i) though you can't use the CLT, by using the normal distribution to estimate the proportion you are assuming that the data is normally distributed.
    hmm I sort of understand.
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    (Original post by Super199)
    hmm I sort of understand.
    Do you have any questions?
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    (Original post by SeanFM)
    Do you have any questions?
    Slightly off tipic but did you do S3 at A-level?
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    (Original post by Super199)
    Slightly off tipic but did you do S3 at A-level?
    S1-S4 but for Edexcel.
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    (Original post by SeanFM)
    S1-S4 but for Edexcel.
    So did you do any mechanics?
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    (Original post by Super199)
    So did you do any mechanics?
    Just M1

    I did C1-C4, FP1-2, S1-4, M1 and D1.
 
 
 
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