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    stuck on probability




    I'm just about done my 2nd assignment and find when I stay at it too long I start to space-out on some of even the more seemingly easier items. I cannot seem to wrap my head around the following even though it looks like it should be easy!:
    compute the 3 probabilities for the following:
    a) at least one 6 when 6 dice are rolled
    b) at least two 6s when 12 dice are rolled
    c) at least three 6s when 18 dice are rolled.
    I'm not sure if I should be using the multiplication or addition rule or if there is some other way of calculating. I think this relates to the idea of complements since it has the "at least one" statement. I wonder if I could just get the hint as to which formula to use? Could this be figured out as a binomial distribution? i.e. X~Binomial(n=6, p=.167)
    E

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    Yes, use the binomial - see my post in the Examples section. Often it is easier to use complementary events when it reduces the amount of computation you have to do.

    a) 1 - P(no 6s)

    b) 1 - [ P(no 6s) + P(one 6) ]

    c) 1 - [ P(no 6s) + P(one 6) + P(two 6s) ]

    Just some advice - when you get tired, put the book away for a while and then come back to it. This stuff is hard enough when you're alert.....

  3. #3
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    binomial et al

    Hi and thanks
    I also tried doing it with pr(notA)5/6^6 which got me .335 and then did 1-.335=.665 giving me a .665 chance of getting at least one 6.

    Can I use the minitab function and/or the binomial distribution table to get the answer or do I have to use the equation? I'll try both but sometimes can't figure out on minitab whether to use pdf, cdf or the inverse calc.

    By the way, is there a glossary of symbols on this website? I'm not totally sure what some of the symbols mean...just learning some of it on my stats calculator. For instance in your example you use a capital C in the equation and I couldn't figure out where that came from.
    Thanks again
    E

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    I also tried doing it with pr(notA)5/6^6 which got me .335 and then did 1-.335=.665 giving me a .665 chance of getting at least one 6.

    --that's fine - the binomial distribution formula will reduce to this anyway

    Can I use the minitab function and/or the binomial distribution table to get the answer or do I have to use the equation? I'll try both but sometimes can't figure out on minitab whether to use pdf, cdf or the inverse calc.

    Sure. You can use any software or the binomial table if you want to - Excel also has a binomial formula that's easy to use.

    By the way, is there a glossary of symbols on this website? I'm not totally sure what some of the symbols mean...just learning some of it on my stats calculator. For instance in your example you use a capital C in the equation and I couldn't figure out where that came from.

    Sorry, no glossary.

    nCr is just shorthand for the combinations formula
    n!/(r!(n-r)!)

    nPr is just shorthand for the permutations formula
    n!/(n-r)!

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    Thankyou, thankyou.
    I'm going to stop now before my brain snaps!
    E

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    Hi John
    Back at it hopefully with a clearer mind. Not sure why I'm getting so stumped onthis question. I looked at your example and we have not covered the permutations or combinations equation yet (using Chance Encounters textbook by Wild & Seber). In just talking to my instructor he is saying to use mintab, use 1.0 as the constant. I believe this is going to get me going. So far I have gotten the following:
    (n=6, p=.167) pr(x=>1) = 1-pr(no 6s)=1-.3341= .6659
    (n=12, p=.167) pr(x=>2) = 1-pr(no 6s)=1-.3802=.6198
    (n=18, p=.167) pr(x=>3) = 1-pr(no 6s)=1-.4012=.5988
    I think this right because the question suggests that a) has the highest probability.
    It is tricky doing this on minitab...Always trying to figure out what the constant is! To do the first question I plugged in 0 as the constant x and n=6,p=.167. For the 2nd I put n=12, p=.167 and x=1. For the 3rd I put n=18, p=.167 and x=2....What do you think?
    E

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    Also...couldn't figure out how to use the binomial distribution table for this one as .167 is not a P value. Do you just round it off? Up or Down if so?
    Thanks
    E

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    You posted:

    (n=6, p=.167) pr(x=>1) = 1-pr(no 6s)=1-.3341= .6659
    (n=12, p=.167) pr(x=>2) = 1-pr(no 6s)=1-.3802=.6198
    (n=18, p=.167) pr(x=>3) = 1-pr(no 6s)=1-.4012=.5988

    but it should be:

    a) 1 - P(no 6s)
    b) 1 - [ P(no 6s) + P(one 6) ]
    c) 1 - [ P(no 6s) + P(one 6) + P(two 6s) ]

    See the difference?

    This is really easy to do in Excel. I haven't used Minitab much, so I can't really help you there.

    Also- not every textbook has a binomial table (haven't seen one in a really long time).

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    Yikes! This is tough. I did see your equation and wasn't sure how it worked.
    In the 2nd equation you have 1 - [ P(no 6s) + P(one 6) ]. My question is what do you base the answer of P(one 6) on? The n=12, p=.167? Or is the P(one 6) based on the first equation?
    E

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    Use the binomial equation with n = 12, r = 1, p = 1/6

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    Okay if I use your equation here is what I get:
    X=>1= 1-P(no 6)= 1-.3341=.6659
    X=>2= 1-(.3341 + .6659)=0 This is where I get stuck because I know the number has to be between 1 and 0
    x=>3= 1-(.3341 + .6559 +.6559)= way over 1.0!!
    I think my problem is partly stemming from trying to get the probability from mintab and not sure if I'm plugginf in the right numbers. I better go see my instructor.
    Thanks
    E

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