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
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.)