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Calculate the chance books will be returned to library

  1. Oct 28, 2008 #1
    Please help probability asap!!:)

    I have tried to understand this problem a million times, and I can not. The answer I have, however it is not good unless I know how to get there and i do not!!!

    A Librarion has estimated that 5% fo the books that people sign out are returned damaged in some way. While 1% of the bookd are never returned at all.

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

    b) if 10 people each sign out ne bool, what is the probability that at least 1 of the books will never be returned?

    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>>??

    the answers ar e

    a) .0135
    b).096
    c).0042

    HELP PLEASE!!!!
     
  2. jcsd
  3. Oct 29, 2008 #2

    CompuChip

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    Re: Please help probability asap!!:)

    a) Let D denote a damaged book being returned, and G a book being returned in a good state.
    1. What is the chance for DDGG, that is: the first two people will return damaged books and the other two don't?
    2. What are the other possibilities? (DGDG, DGGD, ... - you can write them out) What are the probabilities that correspond to them?
    3. What is the chance for 2 D's and 2 G's in any order? Do you see the way to "shortcut" the calculation?

    b) First calculate the chance that they will all be returned (how do you do it? if you don't see the quick way, refer to a). What does this have to do with the actual question?

    Let's do these ones first.
     
  4. Oct 29, 2008 #3
    Re: Please help probability asap!!:)

    Okay I do not understand what the GOod probability is> Like for the first question a) DDGG< the damaged is .05*.05 but is the good then .94? Please help, and with a formula??
     
  5. Oct 30, 2008 #4

    HallsofIvy

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    Re: Please help probability asap!!:)

    If "good" and "bad" are the only two possible outcomes and the probability of "bad" is .05, then the probability of "good" is 1- 0.05= 0.95. I have no idea where you got "0.94".
     
  6. Oct 30, 2008 #5
    Re: Please help probability asap!!:)

    OH!!! haha I jsut re read it!!! I unsderstand!!! OKAY so DDGG is .05*.05*.95*.95 the answer is NOT a!?
     
  7. Oct 30, 2008 #6

    HallsofIvy

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    Re: Please help probability asap!!:)

    Yes, but "two out of four returned damaged" could be DGDG or GGDD or ... (there are 8 possible orders). What is the probability of each of those? What is the total probability?
     
  8. Oct 30, 2008 #7
    Re: Please help probability asap!!:)

    OKay!! That helped me so MUCH!! Thanks !! But here is what i did!! AND it still doesnt work out!!

    So there are 8 ways of doing DDGG, each is .05*.05*.95*.95 correct? Therefore the Probability of each one is .0023. When Multiplying this by 8, the answer is NOT what it shud be!??! So what am I doing wrong???

    AND is there an easier way of finding out there is 8 ways to do so then writing each one out? Is there an equation u can use for that!?

    Thank you again!!!
     
  9. Oct 30, 2008 #8

    HallsofIvy

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    Re: Please help probability asap!!:)

    Yes, the probability of "DDGG" is .05*.05*.95*.95= .00226 and there are 4!/(2!)(2!)= 8 ways of rearranging 4 things, 2 of which are the same and the other 2 the same. Yes, the probability of "two damaged and two good" is 8(.05)(.05)(.95)(.95)= .018. What "shud" the answer be?
     
  10. Oct 30, 2008 #9
    Re: Please help probability asap!!:)

    the answer it has is .0135?? Thats why Im confused!! Cud this be wrong?!
     
  11. Oct 31, 2008 #10

    CompuChip

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    Re: Please help probability asap!!:)

    I think HallsOfIvy has made a typo. (4!)/(2!)(2!) = 6 and not 8. You can write it out:
    DDGG
    DGDG
    DGGD
    GDDG
    GDGD
    GGDD
    and that's all of them.

    Alternative to the reasoning given by HallsOfIvy already, you can say: if I have two good books and two damaged books and four slots to put them into, I can choose 4 slots for the first damaged book, and I have 3 left for the second damaged book. Then the good books automatically go into the remaining two slots. However, I have overcounted by 2 because I can interchange the first two which I placed. This also gives 4 * 3 / 2 = 6.
     
  12. Oct 31, 2008 #11

    HallsofIvy

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    Re: Please help probability asap!!:)

    Oops! Thanks, CompuChip.
     
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