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    Its this June 2009 paper question from OCR.. it uses logs but i swear i have never come across this topic through out the year!


    Q9. repeated independent trials of a certain experiment are carried out. On each trial the probability of success is 0.12.
    i) Find the smallest value of n such that the probability of at least one success in n trials is more than 0.95.

    ii) Find the probability that the 3rd success occurs on the 7th trial

    the answer for part i) says log0.05/log0.88

    i know that 1-0.95 is where they get 0.05 from and 1-0.12 is 0.88

    but i don't understand why/how they used logs and whether logs is always applied with these sorts of questions
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    (Original post by HEY_101)
    Its this June 2009 paper question from OCR.. it uses logs but i swear i have never come across this topic through out the year!


    Q9. repeated independent trials of a certain experiment are carried out. On each trial the probability of success is 0.12.
    i) Find the smallest value of n such that the probability of at least one success in n trials is more than 0.95.

    ii) Find the probability that the 3rd success occurs on the 7th trial

    the answer for part i) says log0.05/log0.88

    i know that 1-0.95 is where they get 0.05 from and 1-0.12 is 0.88

    but i don't understand why/how they used logs and whether logs is always applied with these sorts of questions
    you can use trial and improvement... just keep evaluating (0.88)n until it is less than 0.05.... because you want 1 - 0.88n to be more than 0.95
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    why do you do 0.88n and not use any other numbers

    and how do you use logs for this question, the mark scheme used logs and im confused how they used it
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    oh no wait i get ur method but how do you use logs though??
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    So hard to explain this on here.

    Do you know about Binomial Distribution?

    Let X be the prob of success

    therefore X- B(n, .12)

    We want the P(X>=1) > .95

    P(X>=1) = 1-P(X=0)

    so we want the P(X=0) <= .05

    which is nC0 x .88^n <= .05

    we are left with .88^n <=.05

    you understand it so far? try logging both sides yourself to get the minimum value for n
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    For part two, you know that the P(success) = .12 and P(failure) = .88

    just try working it out by using S for sucess and F for fail and work out how many variations there can be when the 3rd success is on the 7th trial


    e.g SSFFFFS, SFSFFFS, SFFSFFS, etc
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    Quote me for anymore help, feels so great to help people on here after all the help I have got (and hopefully will continue to get) from others on TSR
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    OHHHHHHH yeah i did the working out for log and got the answer finally!!! ill give all u guys reps!!

    and like the title, MANY THANKS!!

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