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    Hello all, I am having a bit of bother with part C of this question - the answer to part A, as you will need it, was 0.3412.

    For part C, I thought I would simply need to do:
    (0.3412)^2 * 0.6588

    I.E. the probability of getting two boxes containing one egg multiplied by the probability of not getting a box containing just one egg. However this not what they did in the markscheme, instead they did something else, yielding a different answer.

    Many thanks, help is appreciated in advance!

    Zacken aymanzayedmannan and other maths wizards to the rescue! <3
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    Binomial expansion assumes statistical independence so P(A and B and C)=P(A)P(B)P(C) But there are 3 boxes. If you use the combinations that can arise

    Let P(D) be the probability when one egg is double yolk and thus D is this event.
    D' is the event not occurring.

    We want 2 with one double yolks and one without. So DDD', this can have different combinations right. We need to times the combinations by the probability of getting DDD'.



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    I still don't really understand
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    (Original post by Cryptokyo)
    Binomial expansion assumes statistical independence so P(A and B and C)=P(A)P(B)P(C) But there are 3 boxes. If you use the combinations that can arise

    Let P(D) be the probability when one egg is double yolk and thus D is this event.
    D' is the event not occurring.

    We want 2 with one double yolks and one without. So DDD', this can have different combinations right. We need to times the combinations by the probability of getting DDD'.



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    So in that sense couldnt I just do three choose two multiplied by (0.3412)^2 and that multiplied by 0.6588?
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    I have changed previously above. And now building on from this. The probability of DDD' is P(D)P(D)P(D'). You can find this but now you need to consider how many combinations of D, D and D' you can have.


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    (Original post by iMacJack)
    So in that sense couldnt I just do three choose two multiplied by (0.3412)^2 and that multiplied by 0.6588?
    But we want combinations and not to choose.


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    Instead we can arrange n objects in n! ways so for 3 objects we can combine them in 6 ways


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    (Original post by Cryptokyo)
    But we want combinations and not to choose.


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    But I'm confused
    This is what they did in the markscheme

    https://gyazo.com/cfe4a1963ea6bf1c8d065e18ae9ffe9a
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    Which paper is this?


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    This is Jan 2003 S2 Cryptokyo
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    Sorry I read the question wrong for part c. The reason for the choosing now is that 3 boxes have been chosen at random.
    Let a = P(D) and b = P(D'). Since we chose 3 at random we obtain a new expansion (a+b)^3 and we are interested in the a^2 b term so it is 3 choose 2


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    (Original post by Cryptokyo)
    Sorry I read the question wrong for part c. The reason for the choosing now is that 3 boxes have been chosen at random.
    Let a = P(D) and b = P(D'. Since we chose 3 at random we obtain a new expansion (a+b)^3 and we are interested in the a^2 b term so it is 3 choose 2


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    Okay mate thank you
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    We use binomial expansion for the 3 boxes as they are independent and have choosing combinations

    Hope this has helped.

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    (Original post by iMacJack)
    ...
    Just read the above exchange and I have no clue what's going on. You're all overcomplicating this. This is a simple case of using the binomial distribution. Let X be the random variable "# of boxes containing exactly 1 egg with a double yolk" then you have 3 boxes, so that's n=3 and the probability of a double yolk is p =0.3412.

    i.e: you have X \sim B(3, 0.412) and you want to find "2 boxes containing exactly 1 egg with a double yolk" so you want \mathbb{P}(X =2).

    Now look in your formula booklet and you should find that \mathbb{P}(X=2) = {3 \choose 2}(0.3412)^2 (1-0.3412)^1.

    This is basically the same thing as you did except you forgot the missing factor of {3 \choose 2}.

    That's because you did 3 boxes: probability box 1 contains * probability box 2 contains * probability box 3 doesn't contain.

    But really, there are three boxes: so you need to add all three probabilities up:

    probability box 1 contains * probability box 2 contains * probability box 3 doesn't contain
    probability box 1 contains * probability box 3 contains * probability box 2 doesn't contain
    probability box 2 contains * probability box 3 contains * probability box 1 doesn't contain

    The binomial distribution was invented precisely to avoid listing all these options down which gets tiresome when you do things with more than just 3 boxes, so use it. You'll get used to spotting it once you do more practice.

    So sum up, you have a binomial distribution with X\sim B(3, 0.3142) and you want \mathbb{P}(X=2).
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    (Original post by Zacken)
    Just read the above exchange and I have no clue what's going on. You're all overcomplicating this. This is a simple case of using the binomial distribution. Let X be the random variable "# of boxes containing exactly 1 egg with a double yolk" then you have 3 boxes, so that's n=3 and the probability of a double yolk is p =0.3412.

    i.e: you have X \sim B(3, 0.412) and you want to find "2 boxes containing exactly 1 egg with a double yolk" so you want \mathbb{P}(X =2).

    Now look in your formula booklet and you should find that \mathbb{P}(X=2) = {3 \choose 2}(0.3412)^2 (1-0.3412)^1.

    This is basically the same thing as you did except you forgot the missing factor of {3 \choose 2}.

    That's because you did 3 boxes: probability box 1 contains * probability box 2 contains * probability box 3 doesn't contain.

    But really, there are three boxes: so you need to add all three probabilities up:

    probability box 1 contains * probability box 2 contains * probability box 3 doesn't contain
    probability box 1 contains * probability box 3 contains * probability box 2 doesn't contain
    probability box 2 contains * probability box 3 contains * probability box 1 doesn't contain

    The binomial distribution was invented precisely to avoid listing all these options down which gets tiresome when you do things with more than just 3 boxes, so use it. You'll get used to spotting it once you do more practice.

    So sum up, you have a binomial distribution with X\sim B(3, 0.3142) and you want \mathbb{P}(X=2).
    Wow. Now that makes a lot more sense! Yeah I wasn't really sure during the above exchange either, this is much more logical though actually, thank you once more Zain
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    (Original post by iMacJack)
    Wow. Now that makes a lot more sense! Yeah I wasn't really sure during the above exchange either, this is much more logical though actually, thank you once more Zain
    No problem. Sorry for the late reply, was watching a movie. :-)
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    (Original post by Zacken)
    No problem. Sorry for the later reply, was watching a movie. :-)
    Don't worry - I'm lucky to have received help anyway haha! Anything good?
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    (Original post by iMacJack)
    Don't worry - I'm lucky to have received help anyway haha! Anything good?
    S'all cool. :yep:
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    (Original post by iMacJack)
    Zacken aymanzayedmannan and other maths wizards to the rescue! <3
    Sorry for the late reply, I woke up like a few hours ago - seems like you were in good hands though!

    (Original post by Zacken)
    No problem. Sorry for the late reply, was watching a movie. :-)
    Go do STEP!!!
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    (Original post by aymanzayedmannan)


    Go do STEP!!!
    How much STEP can I do m8 :rofl:
 
 
 
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