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    I will post some questions. I already tried attempting them, so please just give me the solutions. THANK YOU
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    Post away. *Prays it's not FP3*
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    (Original post by Asad234)
    I will post some questions. I already tried attempting them, so please just give me the solutions. THANK YOU
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    (Original post by Bulletzone)
    Post away. *Prays it's not FP3*
    it is s1 and m2
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    (Original post by Asad234)
    it is s1 and m2
    Hate stats. Will attempt m2.
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    (Original post by Asad234)
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    Part ii) answer is 47/72
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    (Original post by Asad234)
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    Which part, ii)?
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    (Original post by Radioactivedecay)
    Part ii) answer is 25/72
    Some working would be nice. and no
    the following answers according to the textbook is:

    i.2/27
    ii.125/216
    iii.5/9
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    (Original post by Asad234)
    Some working would be nice. and no
    the following answers according to the textbook is:

    i.2/27
    ii.125/216
    iii.5/9
    All dice are fair right?
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    (Original post by thekidwhogames)
    Which part, ii)?
    part i and iii
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    This is A-level? Seems like GCSE. Is this AS?
    Posted on the TSR App. Download from Apple or Google Play
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    (Original post by Bulletzone)
    Hate stats. Will attempt m2.
    Good.
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    Part ii:

    P(6') = 1-P(6) = 5/6.
    Hence 5/6 cubed = 125/216 (as three rolls).
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    (Original post by Asad234)
    Some working would be nice. and no
    the following answers according to the textbook is:

    i.2/27
    ii.125/216
    iii.5/9
    Ahh i realised my error. Part ii is simply 5/6 cubed, which is the probabilty of not obtaining a six when a dice is thrown. Would you like workings fr the other 2 parts?
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    (Original post by Asad234)
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    For part (i), think about the probability of getting all three sixes, and then the probabilities of each of the three ways of getting exactly two sixes; add the results together.

    For part (ii), find the probability of not getting a six with one of the dice, and then cube it.

    For part (iii), I would try to think of all combinations of how you can get at least two numbers the same, and their probabilities together, and then subtract this value from 1.

    I hope that this helps!

    EDIT: Actually, my method for part (iii) is overly long-winded. beckyyfoster did a much better solution three posts down.

    Spoiler:
    Show

    (6/6) * (5/6) * (4/6) = 1 * (20/36) = 5/9
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    (Original post by Radioactivedecay)
    Ahh i realised my error. Part ii is simply 5/6 cubed, which is the probabilty of not obtaining a six when a dice is thrown. Would you like workings fr the other 2 parts?
    yes, please.

    Show me all the working for all the parts please. thank you
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    (Original post by thekidwhogames)
    Part ii:

    P(6' = 1-P(6) = 5/6.
    Hence 5/6 cubed = 125/216 (as three rolls).
    thx
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    okay so:

    ii) probability of no 6's means throwing a 1/2/3/4/5 on all 3 throws
    therefore: 5/6 * 5/6 * 5/6 = (5/6)^3 = 125/216

    iii) different scores on all dice mean throwing any on 1st throw, so then there are only 5 possibilities for 2nd and 4 for third
    therefore: 1 * 5/6 * 4/6 = 20/36 = 5/9
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    (Original post by Asad234)
    yes, please.

    Show me all the working for all the parts please. thank you
    Part i) it is simply the probability of getting two sixes, which can happen 3 different ways plus the probability of getting 3 sixes because it says at least two sixes meaning 2 or greater. So you get 3x (1/6 x1/6 x5/6) + (1/6)^3.
    Part iii is simply 1/6 x 5/6 ( because you have 5 remaining digits to choose from) x4/6 ( because you have 4 remaining digits to choose from). This can happen 6 different ways so we multiply by 6.
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    hh
    (Original post by beckyyfoster)
    okay so:

    ii) probability of no 6's means throwing a 1/2/3/4/5 on all 3 throws
    therefore: 5/6 * 5/6 * 5/6 = (5/6)^3 = 125/216

    iii) different scores on all dice mean throwing any on 1st throw, so then there are only 5 possibilities for 2nd and 4 for third
    therefore: 1 * 5/6 * 4/6 = 20/36 = 5/9
    Thank you very much for the third part
 
 
 
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