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    really really struggling with this topic :confused:

    1-could someone explain why you cant use the tables for geo-distribution unlike binomial???

    2-also, im stuck on a qustion in misc 7 (ocr)
    the probability of success is 1/3
    you continue until a success is made

    what is the probability of a success in ten or fewer attempts?

    3- last lil' bit, in a sample of 486 candidates how many would you expect to have success on attempt 2 (this goes with no2)

    sorrrry for long one!
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    (Original post by VeeOMGlob)
    1-could someone explain why you cant use the tables for geo-distribution unlike binomial???
    You could.

    BUT they don't usually do tables for the geometric distribution. Why?

    Because the terms in a geometric distribution form a geometric progression, for which there is a formula.

    2-also, im stuck on a qustion in misc 7 (ocr)
    the probability of success is 1/3
    you continue until a success is made

    what is the probability of a success in ten or fewer attempts?
    What's the probability of succces on the first attempt? (in terms of p and q)?
    And the second?
    Etc.
    And sum using the formula for the geomtric series.

    3- last lil' bit, in a sample of 486 candidates how many would you expect to have success on attempt 2 (this goes with no2)
    What's the probabilty of one candidate having success on the second attempt?

    And multiply by 486.

    Done.
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    i was just wondering if there was a way to use the binomial tables for questions like this

    its really simple when you put it like that! thanks for taking the time to answer!
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    (Original post by VeeOMGlob)
    i was just wondering if there was a way to use the binomial tables for questions like this

    its really simple when you put it like that! thanks for taking the time to answer!
    Not for the first two parts.

    For the final bit you are in fact using binomial, but you don't need the tables.

    The number of candidates that get it right on the second attempt is \sim B(486,p)

    And the expectation of that distribution is simply 486p.
 
 
 
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