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    probability




    a pharmaceutical firm claims that a certain medication is 92% effective in relieving arthric pain. if ten people who suffer from arthritis are randomly selected and given the medicine, and assuming that the claim is correct, find the probability that

    A) At least eight of them will be relieved of their arthritic pain
    B) Exactly two of them will not be relieved of their arthritic pain

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    I wish we had LaTex. It would be handy for this problem.

    10
    #1. sum(C(10,k)p^kq^(10-k)), p=0.92 and q=0.08
    k=8


    #2. C(10,2)p^2q^8

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    thanks

    thank you for the help. I will try my best.

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    pls help i dont think im on the right track

    A)
    sum(C(10,8) 0.92 ^8 X 0.08^(10-8)), p=0.92 and q=0.08 k=8
    7.36 X 0.64(2)=9.42

    b) C(10,2)0.92^2X0.92 ^8=

    I THINK IM REALLY REALLY LOST.

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    (a)
    use formula for r = 8, 9, 10 and sum them:
    C(10,r) * 0.92^r * 0.08^(10-r)



    (b)
    C(10,2) * 0.08^2 * 0.92^8

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    Thank you. I will try it

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    Checking to see if im correct

    Hi, when i see something like this (10,2) does that mean i should multiply. i dont know what the comma means.

    For question A i got 0.410

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    Checking If Im Correct

    Can someone pls tell me if im correct with this answer. I worked it out again and for

    A) I GOT
    80*7.36*6.4=3768.32

    PLS JUST TELL ME IF IM CORRECT OR NOT

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    We cover these types of problems, using the binomial distribution, in the Examples forum:

    http://www.talkstats.com/showthread.php?t=146

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    Quote Originally Posted by sweetlady
    Can someone pls tell me if im correct with this answer. I worked it out again and for

    A) I GOT
    80*7.36*6.4=3768.32

    PLS JUST TELL ME IF IM CORRECT OR NOT
    NO, you're not correct. You're looking for probability. It must be between 0 and 1. Think about it, 3768.32?,

    C(n,r) means combinations. A selection of r objects from a group of n objects without regard to order. These are the coefficients in your binomial expansion. Given by n!/[(n-r)!r!].

    Since they ask for at least 8, they want 8,9, and 10.

    You have: C(10,8)p^8q^2+C(10,9)p^9q+C(10,10)p^10, where p=0.92 and q=0.08

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    thank you

    thanks john. I will try it.

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    thank you very very much

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    Galactus,
    Take it easy. We're trying to help out beginners here.


    =======================================

    Sweetlady,

    Whenever you see C(n,r) it is short-hand for the formula for the number of combinations of n items, taking r at a time:

    n!/[(n-r)!r!]

    So, for C(10,8), we get: 10! / [ (10 - 8)! * 8! ] which computes to 45

    for C(10,9), we get: 10! / [ (10 - 9)! * 9! ] which computes to 10

    for C(10,10), we get: 10! / [ (10 - 10)! * 10! ] which computes to 1

    So, to compute the probability of "at least 8" we add up the probability of 8, 9, and 10

    P(r) = C(n,r) * p^r * q^(n-r)

    P(8) = C(10,8) * 0.92^8 * 0.08^2 = .1478
    P(9) = C(10,9) * 0.92^9 * 0.08^1 = .3778
    P(10) = C(10,10) * 0.92^10 * 0.08^0 = .4344

    So, P(at least 8) = .1478 + .3778 + .4344 = .96

  14. #14
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    Quote Originally Posted by JohnM
    Galactus,
    Take it easy. We're trying to help out beginners here.

    I'm sorry, I didn't mean to come off as impatient or nasty.

    Note, there were no exclamation points.

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    No worries. Thank you very much for your help

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