librarian

**swoopesjr01** · May 17th 2006, 07:33 AM

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?

**CaptainBlack** · May 17th 2006, 01:14 PM

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

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

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

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 $95\%$ chance of being in the interval $[0.01376,0.01396]$

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:

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

RonL

**ThePerfectHacker** · May 17th 2006, 01:52 PM

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,
$\sum^4_{k=2} {20 \choose k}.05^k.95^{20-k}=.2616$
The other one is,
${20 \choose 2}.05^2.95^{18}=.0159$
Since this is a conjunction probability (this and that)
You need to multiply them to get,
$\approx 0.00415944$

**CaptainBlack** · May 17th 2006, 02:00 PM

Originally Posted by ThePerfectHacker

Again, this is binomial probability thus,
$\sum^4_{k=2} {20 \choose k}.05^k.95^{20-k}=.2616$
The other one is,
${20 \choose 2}.05^2.95^{18}=.0159$
Since this is a conjunction probability (this and that)
You need to multiply them to get,
$\approx 0.00415944$

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

**ThePerfectHacker** · May 17th 2006, 02:03 PM

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.

**CaptainBlack** · May 17th 2006, 08:28 PM

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

**CaptainBlack** · May 18th 2006, 01:47 AM

Originally Posted by ThePerfectHacker

Again, this is binomial probability thus,
$\sum^4_{k=2} {20 \choose k}.05^k.95^{20-k}=.2616$
The other one is,
${20 \choose 2}.05^2.95^{18}=.0159$

Typo here,

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

RonL

**CaptainBlack** · May 18th 2006, 01:48 AM

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

**CaptainBlack** · May 20th 2006, 01:34 PM

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 $0.05/0.99 \approx 0.50505$ ,
and the probability that it is not damages is $1-0.050505 \approx 0.949495$ .

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

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

RonL

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