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AQA Statistics 2 (B) Exam Thursday 21st June

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    (Original post by coolstorybrother)
    So I know that P(a⎪b) = P (a ∩ b) / P (b)
    However in this question 4 cii (below) to my understanding - i may be wrong :L it does not this, to my understanding they are discrete random variables which they have done P (a) / P (b) as shown in the mark scheme
    http://store.aqa.org.uk/qual/gce/pdf...W-QP-JAN11.PDF

    Mark scheme:

    http://store.aqa.org.uk/qual/gce/pdf...W-MS-JAN11.PDF

    However for continuous random variables question 6d they have done :
    P (a ∩ b) / P (b)

    Why is this the case? Is P(a⎪b) for Discrete random variables P (a) / P (b) and for continuous P (a ∩ b) / P (b)? Please explain if i am wrong thanks
    Haha, that's quite hard!

    I don't quite understand your last line, what do you mean?
    The formulas are standard basically; not all of them can be used for every question, it depends on whether or not they are mutually exclusive?

    I.e if they both can happen at the same time or not?

    If they are independent which is the case for continuous random variables -i.e the value can be between any number -like weight of a person can vary between 73kg - 69kg or something then you use P(A n B) = P(A).P(B)

    However, mutually exclusive events such as flipping a coin (you cannot get both heads and tails) are 'discrete' random variables and thus use the other probability formulas available.

    Such as the P(A U B) = P(A) + P(B) - P(A n B)
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    (Original post by OnimushaGTA)
    Haha, I can't wait for this time tomorrow!
    I WILL BE FINISHED! YAYYYYY!


    lol :P
    Lucky you! I still have the pleasure of FP4 on Friday!
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    (Original post by Miller693)
    Lucky you! I still have the pleasure of FP4 on Friday!
    Show those AQA examiners how good you are with your matrices skills!
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    (Original post by OnimushaGTA)
    LALITA!

    Haha! Sometimes that's not always the case though? I think
    Some questions I did, where the value does not exceed the critical value and it was rejected!

    Maybe I misread it :P
    NILESH HAAAY! It's me
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    (Original post by Lollyage)
    SAAME! :yeah:
    I won't be :cry:
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    (Original post by Nerdatious)
    NILESH HAAAY! It's me
    After seeing Sherlock, I quickly realized it's ZOE!!

    Haha, this just proves how much of a nerdy group we are :P
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    (Original post by Lollyage)
    SAAME! :yeah:
    Haha, celebrations are needed!
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    (Original post by coolstorybrother)
    So I know that P(a⎪b) = P (a ∩ b) / P (b)
    However in this question 4 cii (below) to my understanding - i may be wrong :L it does not this, to my understanding they are discrete random variables which they have done P (a) / P (b) as shown in the mark scheme
    http://store.aqa.org.uk/qual/gce/pdf...W-QP-JAN11.PDF

    Mark scheme:

    http://store.aqa.org.uk/qual/gce/pdf...W-MS-JAN11.PDF

    However for continuous random variables question 6d they have done :
    P (a ∩ b) / P (b)

    Why is this the case? Is P(a⎪b) for Discrete random variables P (a) / P (b) and for continuous P (a ∩ b) / P (b)? Please explain if i am wrong thanks
    I think you're getting a little confused. You only need to use P(a⎪b) = P (a ∩ b) / P (b) when the variables are associated (not-independent); if a and b are independent, then P (a ∩ b) = P (a) x P (b) . For c)ii), the easiest way is to consider all the possible combinations; you know x must be greater than 1, and that x + T must be equal to or less than 9, therefore you could have; (2,3), (2,6), (3,3), (3,6), or (4,3). Find the probability of each situation happening, then add (since this is an OR situation; it could be (2,3) OR (2,6) etc, etc).
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    (Original post by PhysicsGirl)
    I think you're getting a little confused. You only need to use P(a⎪b) = P (a ∩ b) / P (b) when the variables are associated (not-independent); if a and b are independent, then P (a ∩ b) = P (a) x P (b) . For c)ii), the easiest way is to consider all the possible combinations; you know x must be greater than 1, and that x + T must be equal to or less than 9, therefore you could have; (2,3), (2,6), (3,3), (3,6), or (4,3). Find the probability of each situation happening, then add (since this is an OR situation; it could be (2,3) OR (2,6) etc, etc).
    You go girl, physics girl!

    (don't know why I said this, sounded good in my head :L)
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    (Original post by OnimushaGTA)
    You go girl, physics girl!

    (don't know why I said this, sounded good in my head :L)
    :rofl:
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    (Original post by OnimushaGTA)
    After seeing Sherlock, I quickly realized it's ZOE!!

    Haha, this just proves how much of a nerdy group we are :P
    :cool:
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    Can anyone please explain to me the logic behind june 2010 question 6- the final final part? I seem to always fail these questions!

    NVM: it just clicked!
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    (Original post by PhysicsGirl)
    I think you're getting a little confused. You only need to use P(a⎪b) = P (a ∩ b) / P (b) when the variables are associated (not-independent); if a and b are independent, then P (a ∩ b) = P (a) x P (b) . For c)ii), the easiest way is to consider all the possible combinations; you know x must be greater than 1, and that x + T must be equal to or less than 9, therefore you could have; (2,3), (2,6), (3,3), (3,6), or (4,3). Find the probability of each situation happening, then add (since this is an OR situation; it could be (2,3) OR (2,6) etc, etc).
    I kind of get what your saying, but then how can you explain what is done in the mark scheme for those questions, because for the same type of question i.e P(a⎪b) ? I'm sure this is stats 1 stuff Just to clarify continuous are independent (every value is random?), and discrete are mutually exclusive




    (Original post by OnimushaGTA)
    Haha, that's quite hard!

    I don't quite understand your last line, what do you mean?
    The formulas are standard basically; not all of them can be used for every question, it depends on whether or not they are mutually exclusive?

    I.e if they both can happen at the same time or not?

    If they are independent which is the case for continuous random variables -i.e the value can be between any number -like weight of a person can vary between 73kg - 69kg or something then you use P(A n B) = P(A).P(B)

    However, mutually exclusive events such as flipping a coin (you cannot get both heads and tails) are 'discrete' random variables and thus use the other probability formulas available.

    Such as the P(A U B) = P(A) + P(B) - P(A n B)
    Sorry if this sounds stupid, but does P(A U B) mean the same as P(a⎪b)? The last line meant :
    Why is this the case? Is P(a⎪b) for Discrete random variables = P (a) / P (b) and for continuous variables = P (a ∩ b) / P (b)? Please explain if i am wrong thanks
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    (Original post by coolstorybrother)
    I kind of get what your saying, but then how can you explain what is done in the mark scheme for those questions, because for the same type of question i.e P(a⎪b) ? I'm sure this is stats 1 stuff Just to clarify continuous are independent (every value is random?), and discrete are mutually exclusive





    Sorry if this sounds stupid, but does P(A U B) mean the same as P(a⎪b)? The last line meant :
    Why is this the case? Is P(a⎪b) for Discrete random variables = P (a) / P (b) and for continuous variables = P (a ∩ b) / P (b)? Please explain if i am wrong thanks
    P( A U B) means the probability of A or B occuring
    P(A|B) means the probability of A occuring given B


    And I think I explained the last line then :L


    The best thing for you to do would be to go and revisit S1 probability laws and how it all works..etc?
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    Can anyone help with june 10 6 biii)
    In the mark scheme they get 0.08 from somewhere.. How?
    Ty
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    (Original post by OnimushaGTA)
    You go girl, physics girl!

    (don't know why I said this, sounded good in my head :L)
    Smooth operator :cool:
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    (Original post by Nerdatious)
    I won't be :cry:
    Nawww!! Think about how relieved you will be this weekend though! Just two more days! :woo:
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    (Original post by OnimushaGTA)
    P( A U B) means the probability of A or B occuring
    P(A|B) means the probability of A occuring given B


    And I think I explained the last line then :L


    The best thing for you to do would be to go and revisit S1 probability laws and how it all works..etc?
    BRAINWAVE!! I get it now, so for the cii, as the events were independent P(A n B)
    was simply multiplying the two probabilities. Whereas with the continuous random variables as the are not associative, you have to use the other formula the P(A|B) one :L Could you confirm thats right please?
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    (Original post by coolstorybrother)
    BRAINWAVE!! I get it now, so for the cii, as the events were independent P(A n B)
    was simply multiplying the two probabilities. Whereas with the continuous random variables as the are not associative, you have to use the other formula the P(A|B) one :L Could you confirm thats right please?
    Spot on!
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    (Original post by In One Ear)
    Can anyone please explain to me the logic behind june 2010 question 6- the final final part? I seem to always fail these questions!

    NVM: it just clicked!
    HOw do you do it! Explain plz! Where do they get there numbers from!

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