Question 535145
Assuming the chance of a misprint being on any page is uniform and that each 

misprint is an independent event, we can think of a misprint being randomly 

assorted into 600 pages and this occurrence happening 200 times. Therefore, the 

probability a misprint will go to any page is 1/600 since there are 600 pages and 

a misprint may go into any of them evenly. Then, like flipping coins, the next 

misprint event happens independently of the previous one and so its probability of 

going to any page is also 1/600, and so on and so on for all 200 misprints. Then, 

just like coins, you can multiply their probabilites together. For example, what 

is the probability of flipping 3 Heads in 3 coin tosses: Simply (1/2)*(1/2)*(1/2). 

The same logic can be used here: For instance, the probability that one page will 

have all 200 misprints is (1/600)^200.
 Part (a) asks that there are exactly 2 misprints on one page and the other 198 

misprints must be on the other 599 pages. So you can do 
(1/600) * (1/600) * (599/600)^198 * 200C2
where 200C2 is 200 Choose 2 or C(200,2) [Combinations: I'll explain later]

Part (b) asks that one page contains 2 or more misprints. We could calculate the 

probability of a page getting exactly 2 misprints and then add to it the 

probability of getting exactly 3 misprints and then add the probability of getting 

exactly 4 misprints and so on, but that is time consuming so we can just subtract 

the probability of getting exactly one misprint on that page and exactly no misprints on  

that page from 1, since this is the same probability. So

1 - (1/600) * (599/600)^199 * 200C1 - (599/600)^200 * 200C0

To convince yourself of this logic, use smaller numbers and work with actual 

paper. For example, suppose you have only 6 pages and 2 misprints going among 

them. A misprint may go to the 6 pages evenly. So the chance page 1 will have 2 

misprints is :

(1/6) * (1/6)
The first misprint must go to page 1 AND the second misprint must go to page 1.

exactly one misprint:

(1/6) * (5/6) + (5/6) * (1/6)
The first misprint must go to page 1 AND the second misprint must go to pages 2-6
OR
The first misprint must go to pages 2-6 AND the second misprint must go to page 1

no misprints:

(5/6) * (5/6)
The first misprint must go to pages 2-6 AND the second misprint must go to pages 2-6.

one or more misprints:
(1/6) * (5/6) + (5/6) * (1/6) + (1/6) * (1/6) = 1- (5/6) * (5/6)
Either write the case for exactly one misprint plus the case for exactly two 

misprints or subtract the case for exactly no misprints from one. 

Even though I write first and second misprint, they are independent events so the order of 

the misprints is inconsequential but to keep track of which misprint goes where, I 

order them.

Now imagine you have four misprints in 6 pages.

P(Exactly 4 misprints on a given page) =
(1/6) * (1/6) * (1/6) * (1/6)

P(Exactly 3 misprints on a give page) =
(1/6)*(1/6)*(1/6)*(5/6) + (1/6)*(1/6)*(5/6)*(1/6) + (1/6)*(5/6)*(1/6)*(1/6) +
(5/6)*(1/6)*(1/6)*(1/6) = (1/6)*(1/6)*(1/6)*(5/6)*4

P(Exactly 2 misprints on a given page) =
(1/6)*(1/6)*(5/6)*(5/6) + (1/6)*(5/6)*(1/6)*(5/6) + (1/6)*(5/6)*(5/6)*(1/6) + 
(5/6)*(1/6)*(1/6)*(5/6) + (5/6)*(1/6)*(5/6)*(1/6) + (5/6)*(5/6)*(1/6)*(1/6) =
6 * (1/6)*(1/6)*(5/6)*(5/6)

As you can see, the number being multiplied is just the differnt combinations of 

misprints appearing from the total number of misprints. So nCr = (# of total 

misprints) C (# of misprints appearing on target page). For the case of exactly 2 

misprints on a given page that preceded, this is 4C2 or 6.