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Thread: Please help!! Basic Probability.

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    Angry Please help!! Basic Probability.




    Hi im having trouble with this question.
    If anyone could help me out thatd be great , Im not sure where to start!


    Heres info:

    The cost of a phone call passed on to a 'live' operator is approx. ten times that of a call answered by an automated customer-service system.However , as more and more companies have implemented automated-systems, customer annoyance with these sytems have grown.Many customers are quick to leave the automatedsystem when given an option such as ''press 0 to talk to a live operator" . Research has shown that approxiamtely 40% of all callers to automated systems will automatically opt to go to a live operator when given the chance.

    And questions:


    What is the probability that, from a total of 10 callers, none will automatically opt to talk to a live operator?




    What is the probability that , from a total of ten callers, at least 5 will automatically opt to talk to a live operator?




    What assumptions are necessary for the validity of the two above answers?

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    Just so you know Im not sure where to start working this problems out , if someone could point me in the right direction to start with.?

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    ok so far ive worked out:

    Probability that out of ten callers none will opt for live operator - 0.6

    Probability that out of ten callers at least 5 will opt for a live operator - 0.4

    stuck on last part.i thought it might have somthing to do with the call price or maybe that all calls are same in some way? or to same company? not sure

    Are those answers correct? or close? am i even warm?

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    Quote Originally Posted by el1 View Post
    ok so far ive worked out:

    Probability that out of ten callers none will opt for live operator - 0.6

    Probability that out of ten callers at least 5 will opt for a live operator - 0.4

    stuck on last part.i thought it might have somthing to do with the call price or maybe that all calls are same in some way? or to same company? not sure

    Are those answers correct? or close? am i even warm?
    No they are not.

    Have you been studying the binomial distribution?

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    yeah i had a thought thats what is was. Im a bit confised by it to be honest.
    How to i use to given info in the formula?

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    you have a total number of n=10 callers

    the probability they opt for an operator if p=.4

    so you have a Bin(10,.4) distribution

    Let X~Bin(10,.4)

    compute

    P(X=0)

    and

    P(X>=5)

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    is that using excel? im not to familiar with excel i prefer to do things on paper, is that an issue , is this too complicated that i must use excel?

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    you either use the binomial distribution (using a calculator) or you can use a binomial table...there should be one in your book.

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    i have the binomial formulae here , im confused as to what n! means?

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    Quote Originally Posted by el1 View Post
    i have the binomial formulae here , im confused as to what n! means?
    n!=n*(n-1)*(n-2)*....*3*2*1

    ex.

    5!=5*4*3*2*1=120

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    so if n=10 then n! equals

    n*(10-1)*(10-2)*(10-3).....*(10-10)

    is that right?

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    Quote Originally Posted by el1 View Post
    so if n=10 then n! equals

    n*(10-1)*(10-2)*(10-3).....*(10-10)

    is that right?
    10!=10*9*8*7*6*5*4*3*2*1=3628800

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    i mean

    n=10

    =10*9*8*7*6*5*4*3*2*1

    ?

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    you can check your final answer here...


    http://www.stat.tamu.edu/~west/apple...omialdemo.html

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    and the bottom part of the formula says


    X!(n-x)! which if x=0 then


    o*10*9*8.... is that right?

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