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Math Help - Binomial distribution

  1. #1
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    Unhappy Binomial distribution

    I would like to knw why dont we treat getting a 4,5,6 as a success and treat getting a 1,2,3 as failure as a binomial problem,two outcomes why then in ex7.7.2 a) file attached says not a binomial model
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    Last edited by spacemenon; July 9th 2013 at 07:05 AM. Reason: file not atttached
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  2. #2
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    Re: Binomial distribution

    You cannot say that getting 4,5,6 is a success because getting ten 4s is a fail.
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  3. #3
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    Re: Binomial distribution

    To be considered a binomial experiment, each event has to be able to be classified as one of two states (call them "success" and "failure"). Each of the 10 rolls is an event, and the face that has been rolled needs to be classified as a success or failure. For instance, an even number and an odd number. This is not the case here, where we are looking for a range, or multiple number of outcomes. In fact, this problem is a multinomial experiment.
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  4. #4
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    Re: Binomial distribution

    I don't understand Shakarri's response. Yes, getting 10 4s is a fail but the question is about getting three 4s, four 5s, and three 6s in 10 rolls of a fair die.

    spacemenon, we are NOT "counting 4, 5, and 6 as a success". Were counting, specifically, three 4s, four 5s, and three 6s as a success because that is what the problem asks about. I don't know what you mean by "treat getting a 1,2,3 as failure" because I see no mention of "1, 2, 3" on your post.

    What that is saying is that a binomial distribution always involves trials in which there are two possible outcomes. Because the possible events are 1, 2, 3, 4, 5, 6 this is NOT binomial.

    (The simple fact that we are using a die, with 6 possible outcomes does not itself mean that any "event" based on it cannot be binomial. If we ask about 'even' or 'odd' we now have two events so this is binomial. If we ask "greater than or equal to 3 or less than 3" we have two events and so binomial. if we ask "evenly divides 6 or not" we have two events ("1, 2, 3, and 6" and "4, 5') and so binomial
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    Re: Binomial distribution

    that means if there was a special die where i would get three 4s OR four 5s OR three 6s in one throw(strictly imaginery) then i could CLUB ALL THE THREE EVENTS call three 4s, four 5s, and three 6s as a success and otherwise a failure and a binomial distribution,or

    normal dice i could club 4,5,6 has success since in each trial i could get 4 or 5 or 6 but in the above case ( in the problem)for each trial u will never get three 4s OR four 5s OR three 6s when u throw a dice once ,

    IS MY OBSERVATION RIGHT pls comment
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    Re: Binomial distribution

    pls comment, if thirteen cards are picked frm ordinary pack ,then find the probabilty of getting 4 clubs,4 diamonds,3 hearts and 2 spades ,is it a good binomial model?f


    here spade or diamond or heart or club as success for each trial here has 4 outcomes , is this the reason why we r not considering the above as this as a binomial model
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    Re: Binomial distribution

    Quote Originally Posted by spacemenon View Post
    pls comment, if thirteen cards are picked frm ordinary pack ,then find the probabilty of getting 4 clubs,4 diamonds,3 hearts and 2 spades ,is it a good binomial model?f


    here spade or diamond or heart or club as success for each trial here has 4 outcomes , is this the reason why we r not considering the above as this as a binomial model
    You should use a multinomial model

    Quote Originally Posted by HallsofIvy View Post
    I don't understand Shakarri's response. Yes, getting 10 4s is a fail but the question is about getting three 4s, four 5s, and three 6s in 10 rolls of a fair die.
    His idea was to make 4, 5 and 6 all successes and 1, 2 or 3 all fails; I gave a counterexample which showed that getting 10 successes didn't lead to the overall success which he was aiming for.
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  8. #8
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    Re: Binomial distribution

    Let me add: If we are asking about either getting "three 4s, four 5s, and three 6s in 10 rolls of a fair die" or anything else, then we are considering only two possible outcomes and so that would be a binomial distribution.
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