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using binomial distribution watch

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    i do not get this q.
    it is do to with which is most likely rolling one 6 in 6 throws or rolling two sixes in 12 throws,
    On a single throw of a die the probability of getting a 6 is a 1/6 and that of not getting a 6 is 5/6.
    So the probability distributions for the 2 situations in the problem are B(6,1/6) and B(12,1/6) giving probabilities of:

    1-6C0 (5/6)^6 = 1-0.335=0.665 (at least one 6 in 6 throws)

    and

    1-[12C0(5/6)^12+12C1(5/6)^11(1/6)]=1-(0.112+0.269)
    =0.619 (at least 2 sixes in 12 throws)

    So at least one 6 in 6 throws is more likely.

    what i dont get is why in 6 throws^^ the r value is 0 if they need one 6 as usually P(X=R) in this case i would of thought that P(X=1) instead of P(X=0)
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    can ne 1 help plz
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    (Original post by kam130)
    i do not get this q.
    it is do to with which is most likely rolling one 6 in 6 throws or rolling two sixes in 12 throws,
    On a single throw of a die the probability of getting a 6 is a 1/6 and that of not getting a 6 is 5/6.
    So the probability distributions for the 2 situations in the problem are B(6,1/6) and B(12,1/6) giving probabilities of:

    1-6C0 (5/6)^6 = 1-0.335=0.665 (at least one 6 in 6 throws)

    and

    1-[12C0(5/6)^12+12C1(5/6)^11(1/6)]=1-(0.112+0.269)
    =0.619 (at least 2 sixes in 12 throws)

    So at least one 6 in 6 throws is more likely.

    what i dont get is why in 6 throws^^ the r value is 0 if they need one 6 as usually P(X=R) in this case i would of thought that P(X=1) instead of P(X=0)
    are you sure its P(X=0) cuz that doesn't make sense! Maybe its because the prob of at least 1 is the prob of 1 - P(X=0)... is that it? :confused:
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    na i jus dont get it wot exam board u doin
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    (Original post by kam130)
    na i jus dont get it wot exam board u doin
    OCR/MEI... is this S2 for you? if so do you have your exam this friday?
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    na s1, if ya got dat book chek out top of page 149
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    by the way if i was taclking this question i wouldn't have gone about it that way... as the question doesn' ask you to find "at least" i would do it like this:

    for the prob of 1 in 6 throws:

    6C1.(1/6)^1.(5/6)^5 = 0.402

    for the prob of 2 in 12 throws:

    12C2.(1/6)^2.(5/6)^10 = 0.296

    therefore there is a higher probability of 1 in 6 throws...
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    (Original post by kam130)
    na s1, if ya got dat book chek out top of page 149
    which board?
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    same as u, chek out page 149 if ya got it, (mei, ocr)
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    iv had a look in the s1 book i have and the pages only go up to 140 so i must have the old version... what section is it in? probability or binomial dis?
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    binomial distribution
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    ok no i don't have that question in my book... is the answer i did above not the correct one then? if not try typing the exact question out and il have a look
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    the exact q is at d start of dis thread, but i dont understand, y P(X=0) when it should be P(X=1) i think newayz, btw wot dya get in S1
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    but where in the question does it specify that P(X=0)??
    I got an A in s1
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    it doesnt 6C0 tells me that r=0 and P(X=0) where R=0
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    but theres no need to use 6C0 in the question... you would need to if the question was to find the probability of "at least" 1 six in 6 throws in which case the prob is 1 - P(X=0) . If this is what the question asks then it would be:

    for at least 1 six in 6 throws:

    6C0.(1/6)^0.(5/6)^6 = 0.335
    1 - 0.335 = 0.665
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    which is what you'v written so i don't see where the problem is...
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    The question is asking for the probability of AT LEAST 1 Six.

    P(X=1) is the probability of EXACTLY 1 Six. The question DOES NOT ASK THIS.

    You are asked to calculate p(X>0) = 1 - P(X=0)

    The probability of P(X=0) is just (5/6)^6 - you can multiply by 6c0 if you want to, but that just equals 1 anyway, so it's not required (because there is only one combination of not getting a six 6 times)


    Likewise in question 2, you are asked for the p(X>1) = 1 - P(X=0) - P(X=1)

    P(X=0) is easy....just (5/6)^12 again (the 12c0 is not required)

    P(X=1) is just (5/6)^11 * (1/6), because that is the probability of getting one 6 in 12 throws. BUT THEN you must multiply by 12C1 (which equals 12), because there are 12 combinations of getting just one six (you could throw the 6 in your first throw, or in your second, or third, or fourth...and so on - which equals 12 different possiblities)

    Hope that's clear
 
 
 
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