1. ## Probability

Hi, can someone help me with this probability problem?

A publisher is printing a 1200 page book. The machinery is good but not perfect, printing an unintended blotch on average once every 100 pages. A book will be rejected if there are 20 or more pages with a blotch. In a production run of 10,000, about how many books do you expect will be rejected?

I was told the answer is 102 (probability 0.0102 of a book's rejection) but i just can't seem to work it out. Your help is ver appreciated, and thank you in advance.

2. Originally Posted by ayato
Hi, can someone help me with this probability problem?

A publisher is printing a 1200 page book. The machinery is good but not perfect, printing an unintended blotch on average once every 100 pages. A book will be rejected if there are 20 or more pages with a blotch. In a production run of 10,000, about how many books do you expect will be rejected?

I was told the answer is 102 (probability 0.0102 of a book's rejection) but i just can't seem to work it out. Your help is ver appreciated, and thank you in advance.
The probability that a page is blotched is 0.01. There are 1200 pages in a
book, so the number of blotched pages has a binomial distribution:

P(n blotch)=B(n, 0.01, 1200)=1200!/(n!(1200-n)!) 0.01^n 0.99^(1200-n)

Now we could sum the appropriate number of terms of this,
(or rather use P(n>=20)=1-P(n<=19)=1-sum(n=0..19) P(n))
but I expect you are supposed to use the normal approximation
to the binomial at this point.

The normal approximation has a m= 0.01*1200=12 blotches, and a standard
deviation s=sqrt(1200*0.01*0.99)~=3.45.

In the normal approximation because it is continuous we translate 20 or
more as 19.5 or more, then we are asking for the probability of a normal
random variable of mean 12 and sd ~=3.45 exceeding 19.5.

(this approximation gives a probability of ~0.0148, rather than 0.102, so
I guess you are supposed to use the exact binomial probability.)

RonL