Probability of a page being free from printing errors

[shakgoku]

A book of 100 pages contains 200printing errors distributed randomly among the pages. the probability that one of the pages will be completely error free is closest to
a)67%
b)50%
c)25%
d)13%

The Attempt at a Solution

I tried to calculate total number of possible events. I got struck there as it was getting too complicated.
Ex all 200errors can be in 1 page in 100 ways
Next suppose we distribute errors on any 2pages only then we have to find number of pairs of pages possible which might be 100+99+98+...+1(counted by me for smaller numbers) multiplied by 199
(ex- 1+199,2+198,3+197... Upto 199+1) after that I needed to find groups of 3. And it was too hard and complicated.then i realized that With this approach i might count same event more than once. So i stopped.
There should be a easier way.

 

[lanedance]

how about trying to find the complement first, what is the probability that every page has at least one error?

also looking at "combinations with replacement" could be useful for this problem

 

[Yayness]

I'm assuming you mean the probability for a randomly chosen page being completely error free.

Hint: What is the probability that a certain error will not be on the page?

 

[lanedance]

i would probably read it as at least one page is error free, but it is open for interpretation & that will be easier to solve

 

[Reyzal]

I came out to 13.4% chance of not getting an error on the First Page, (84.6% chance of getting an error on first page)
then 1-.846^100=99.99% chance of having at least 1 page with no error?

Can someone tell me if I am completely wrong? I feel like my math was solid but still not confident in my answer (cause I thought it wouldn't be nearly that high)

 

[Yayness]

i would probably read it as at least one page is error free, but it is open for interpretation & that will be easier to solve

Lanedance, you are right. It wasn't clear what the problem was asking for, but I'm still pretty sure it was meant to be the probability that a randomly chosen page will be error free, as that is closest to one of the alternatives.

I didn't find any easy ways to get the probability for at least one page being error free, or the probability for exactly one page being error free. Hopefully, these calculations are right:
The probability that at least one page is error free:
\frac{\sum^{99}_{k=1}\binom{100}{k}k^{200}(-1)^{k-1}}{100^{200}}\approx 0.99999997868=100\%

The probability that exactly one page is error free:
\frac{100\sum^{99}_{k=1}\binom{99}{k}k^{200}(-1)^{k-1}}{100^{200}}\approx 5.31379\cdot 10^{-7}=0\%

Don't worry! The probability for a randomly chosen page being error free, is a lot easier to find. At least you don't need to calculate the sum of 99 different terms. I will not give the solution in this post, but this hint could be useful:
Choose one page and one error. What is the probability that this error will not be on the page? How is that useful when you want to find the probability that 200 errors will not be on the page?