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    (Original post by Someboady)
    Yes thank you!

    The mark scheme for the second one shows:

    P(2 weigh more than 53kg and 1 less) = 3 x 0.0668^2 x (1 - 0.0668)
    Oh sorry, in answer to another question - yes. Anything on a venn diagram is either going to be an intersection or a probability by itself. You won't find 'A given B' directly as far as I am aware.

    Ah yes, I forgot -there are basically three ways of arranging it. It's more of an S2 thing to be honest, so it is awfully mean to chuck it in there. I've explained it in some detail on a few threads here but can't find any of them.

    Basically, imagine a more simple example - tossing a coin.

    If you toss it twice, then there are two ways of getting a head and a tails - HT and TH. There is only 1 way of getting 2 tails and 2 heads, TT and TH so if I asked you what the probability of finding a head and at ails was, then you would work out the probability of getting a head and tails (0.5 * 0.5) and then multiply it by 2 because it can either be HT or TH.

    If you toss it three times, however, then in a similar way, 3 heads is just HHH. 2 heads / 1 tail is just HHT, HTH or THH = 3 * P(HHT). 1 head / 2 tails is HTT, THT or TTH = 3 * P(HTT) and 3 tails is TTT. So in that example - it's just like finding HHT - there are 3 ways of doing so so you find the probability of finding it once and multiply it by the number of ways possible.

    There is a general formula, which is number of ways = \frac{n!}{(k_1)!(k_2)!(k_3)!...} where n is the total number of objects/trials (in this case 3), and the k's represent the different types of objects and their numbers. In the coin example, there would only be k1 and k2, where k1 is heads and k2 is tails. In more complicated examples you can use more k's where there are more than 2 types of objects. oh, and \sum_{i=1}^n k_i = n.

    So in the 3 tosses example, k1 = number of heads = 2 and k2 = number of tails = 1 and n = number of tosses = 3 and so the number of permutations is \frac {3!}{2!\times1!} = 3.
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    (Original post by SeanFM)
    Oh sorry, in answer to another question - yes. Anything on a venn diagram is either going to be an intersection or a probability by itself. You won't find 'A given B' directly as far as I am aware.

    Ah yes, I forgot -there are basically three ways of arranging it. It's more of an S2 thing to be honest, so it is awfully mean to chuck it in there. I've explained it in some detail on a few threads here but can't find any of them.

    Basically, imagine a more simple example - tossing a coin.

    If you toss it twice, then there are two ways of getting a head and a tails - HT and TH. There is only 1 way of getting 2 tails and 2 heads, TT and TH so if I asked you what the probability of finding a head and at ails was, then you would work out the probability of getting a head and tails (0.5 * 0.5) and then multiply it by 2 because it can either be HT or TH.

    If you toss it three times, however, then in a similar way, 3 heads is just HHH. 2 heads / 1 tail is just HHT, HTH or THH = 3 * P(HHT). 1 head / 2 tails is HTT, THT or TTH = 3 * P(HTT) and 3 tails is TTT. So in that example - it's just like finding HHT - there are 3 ways of doing so so you find the probability of finding it once and multiply it by the number of ways possible.

    There is a general formula, which is number of ways = \frac{n!}{(k_1)!(k_2)!(k_3)!...} where n is the total number of objects/trials (in this case 3), and the k's represent the different types of objects and their numbers. In the coin example, there would only be k1 and k2, where k1 is heads and k2 is tails. In more complicated examples you can use more k's where there are more than 2 types of objects. oh, and \sum_{i=1}^n k_i = n.

    So in the 3 tosses example, k1 = number of heads = 2 and k2 = number of tails = 1 and n = number of tosses = 3 and so the number of permutations is \frac {3!}{2!\times1!} = 3.
    This was very useful! Thanks!
    Unfortunately I think I've done nearly every question on Probability and I still don't feel confident doing question :/ Maybe I rush the questions too much or something...The exams in 3 days or so... What should I do now ? :L
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    (Original post by Someboady)
    This was very useful! Thanks!
    Unfortunately I think I've done nearly every question on Probability and I still don't feel confident doing question :/ Maybe I rush the questions too much or something...The exams in 3 days or so... What should I do now ? :L
    Find some more questions in the textbook or online, I guess (madasmathas/solomon papers). If probability is the only placce you're struggling with then don't worry too much, just get everything else right and attempt probability as much as you can.. you'll probably get some marks
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    (Original post by SeanFM)
    Find some more questions in the textbook or online, I guess (madasmathas/solomon papers). If probability is the only placce you're struggling with then don't worry too much, just get everything else right and attempt probability as much as you can.. you'll probably get some marks
    Yes, I'll try my best thanks! I'm hoping for full UMS so I need to smash probability in the exam xD. Lets hope its an easy question and I don't make too many silly mistakes.
    Also just for clarification...
    With probabilty do we give our answers as a decimal to 3sf or leave it as a fraction?
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    (Original post by Someboady)
    Yes, I'll try my best thanks! I'm hoping for full UMS so I need to smash probability in the exam xD. Lets hope its an easy question and I don't make too many silly mistakes.
    Also just for clarification...
    With probabilty do we give our answers as a decimal to 3sf or leave it as a fraction?
    I'm not sure have a look at past papers and see what they say.
 
 
 
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