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  1. #1
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    librarian

    A librarian has estimated that 5% of the books that people sign out are resturned damaged in some way, while 1% of the books are never returned at all.

    a) if 4 people each sign out one book and all are returned, what is the probability that exactly 2 of the books will be returned damaged? 0.0135

    b) if 10 people each sign out one book, what is the probability that between 2 and 4 (inclusively) are returned damaged and exactly 2 are never returned at all? 0.096

    c) If 20 people each sign out one book, what is the probability that between 2 and 4 (inclusively) are returned damaged and exactly 2 are never returned at all? 0.0042

    any idea?
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  2. #2
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    Quote Originally Posted by swoopesjr01
    A librarian has estimated that 5% of the books that people sign out are resturned damaged in some way, while 1% of the books are never returned at all.

    a) if 4 people each sign out one book and all are returned, what is the probability that exactly 2 of the books will be returned damaged? 0.0135
    Binomial distribution problem.

    Out of every $\displaystyle 100$ books signed out $\displaystyle 5$ are returned damaged and $\displaystyle 1$ is not
    returned at all. So of every $\displaystyle 99$ books returned $\displaystyle 5$ are damaged so the probability
    that a returned book is damaged is $\displaystyle 5/99\approx 0.0505$

    Of $\displaystyle 4 $ books returned the probability that exactly $\displaystyle 2$ are damaged is:

    $\displaystyle P={4 \choose 2} 0.0505^2 (1-0.0505)^2$$\displaystyle =\frac{4!}{2!\ 2!} 0.0505^2\ 0.9495^2\approx 0.01379$

    Ooops - the above is the probability that of four returns two are damaged
    irrespective of how many were signed out.
    - No in retrospect this is right
    I just managed to confuse myself when rereading the explanation as it is not
    clear enough. There is a clearer (at least I think so) explanation in one of my
    later posts.


    Simulation shows that the proper probability has about $\displaystyle 95\%$ chance of being in the interval $\displaystyle [0.01376,0.01396]$


    Now if the question had been:

    a) if 4 people each sign out one book, what is the probability that exactly 2
    of the books will be returned damaged?

    Then the answer would be:

    $\displaystyle P_1={4 \choose 2} 0.05^2 (1-0.05)^2=\frac{4!}{2!\ 2!} 0.05^2\ 0.95^2\approx 0.0135$

    RonL
    Last edited by CaptainBlack; May 20th 2006 at 01:40 PM. Reason: Correction to simulation results
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  3. #3
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    Quote Originally Posted by swoopesjr01

    c) If 20 people each sign out one book, what is the probability that between 2 and 4 (inclusively) are returned damaged and exactly 2 are never returned at all? 0.0042
    Again, this is binomial probability thus,
    $\displaystyle \sum^4_{k=2} {20 \choose k}.05^k.95^{20-k}=.2616$
    The other one is,
    $\displaystyle {20 \choose 2}.05^2.95^{18}=.0159$
    Since this is a conjunction probability (this and that)
    You need to multiply them to get,
    $\displaystyle \approx 0.00415944$
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    Quote Originally Posted by ThePerfectHacker
    Again, this is binomial probability thus,
    $\displaystyle \sum^4_{k=2} {20 \choose k}.05^k.95^{20-k}=.2616$
    The other one is,
    $\displaystyle {20 \choose 2}.05^2.95^{18}=.0159$
    Since this is a conjunction probability (this and that)
    You need to multiply them to get,
    $\displaystyle \approx 0.00415944$
    The trouble is that the two clauses in the conjunction are not independent -
    but multiplying may well be OK-ish here but I would like to be more confident
    than I am.

    RonL
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    Quote Originally Posted by CaptainBlack
    The trouble is that the two clauses in the conjunction are not independent -
    but multiplying may well be OK-ish here but I would like to be more confident
    than I am.

    RonL
    I was troubled by that too, I just saw that my hypothesis matched his answers. So I ignored that.
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    Quote Originally Posted by ThePerfectHacker
    I was troubled by that too, I just saw that my hypothesis matched his answers. So I ignored that.
    I was distrubed by the given answers to all three questions in the original
    post, that is why I stopped after doing (a) - I had to go away and think
    about what was going on.

    If I get the chance (and I probably will as my research budget has been
    frozen due to contract delays) I plan to see if simulation will throw any light
    on the correctness of the given answers.

    (Rule 1 of Applied Probability: When in doubt about the correctness of a
    probability calculation - simulate)

    RonL
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    Quote Originally Posted by ThePerfectHacker
    Again, this is binomial probability thus,
    $\displaystyle \sum^4_{k=2} {20 \choose k}.05^k.95^{20-k}=.2616$
    The other one is,
    $\displaystyle {20 \choose 2}.05^2.95^{18}=.0159$
    Typo here,

    $\displaystyle {20 \choose 2}0.01^2\ 0.99^{18}=.0159$

    RonL
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  8. #8
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    Quote Originally Posted by CaptainBlack
    I was distrubed by the given answers to all three questions in the original
    post, that is why I stopped after doing (a) - I had to go away and think
    about what was going on.

    If I get the chance (and I probably will as my research budget has been
    frozen due to contract delays) I plan to see if simulation will throw any light
    on the correctness of the given answers.

    (Rule 1 of Applied Probability: When in doubt about the correctness of a
    probability calculation - simulate)

    RonL
    Simulations done.

    For b) they give p~0.00023

    for c) they give p~0.00357

    RonL
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  9. #9
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    Quote Originally Posted by swoopesjr01
    A librarian has estimated that 5% of the books that people sign out are resturned damaged in some way, while 1% of the books are never returned at all.

    a) if 4 people each sign out one book and all are returned, what is the probability that exactly 2 of the books will be returned damaged? 0.0135
    Given that a book is returned the probability that it is damaged is $\displaystyle 0.05/0.99 \approx 0.50505$,
    and the probability that it is not damages is $\displaystyle 1-0.050505 \approx 0.949495$.

    So the probability that if four books are returned from four checked out, that
    exactly two are returned damaged is:

    $\displaystyle
    P_1={4 \choose 2} 0.050505^2 (1-0.050505)^2=$$\displaystyle \frac{4!}{2!\ 2!} 0.050505^2\ 0.949495^2\approx 0.013798
    $

    RonL
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