# librarian

• May 17th 2006, 07:33 AM
swoopesjr01
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?
• May 17th 2006, 01:14 PM
CaptainBlack
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. :mad: 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 • May 17th 2006, 01:52 PM ThePerfectHacker 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 • May 17th 2006, 02:00 PM CaptainBlack 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 • May 17th 2006, 02:03 PM ThePerfectHacker 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. • May 17th 2006, 08:28 PM CaptainBlack 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 • May 18th 2006, 01:47 AM CaptainBlack 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 • May 18th 2006, 01:48 AM CaptainBlack 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 • May 20th 2006, 01:34 PM CaptainBlack 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