Probability Question (stats)

Question

A book has 500 pages on which typographical errors could occur. Suppose that there are exactly 10 such errors randomly located on those pages.

$a)$ Find the probability that a random selection of 50 pages will contain no errors.

$b)$ Find the probability that 50 randomly selected pages will contain at least 2 errors. $$\\$$ This is what I have so far:

$a)$ $N$= 500 $n$=50 $p$= $\frac{10}{500}$

$\frac {\binom{10}{0} \binom{500-10}{50}}{\binom{500}{50}}$

Answer 1

Here we assume each page could have at most one of the ten errors, and thus they are not independent. Notice that each page is different, and thus we use ordered arrangement/permutation instead of combination (i.e. order matters).

For a) $$\frac{A_{490}^{50}}{A_{500}^{50}}$$ For the above, the whole space has permutation of $A_{500}^{50}$. For the case where all selections are non-error, we have $490$ pages in total, and we select $50$ pages out - which is $A_{490}^{50}$. Notice that we are not doing $\binom {490}{50}$, because each page is different (they are labels as page 1, page 2, etc.), and thus the case of having error on page 1 v.s. having error on page 25 are different cases. The analysis is similar for b).

For b), you exclude cases when there is no error pages and one error page. $$1-\frac{A_{490}^{50}}{A_{500}^{50}}-\frac{A_{10}^{1}A_{490}^{49}}{A_{500}^{50}}$$

Where $A_n^r=\frac{n!}{(n-r)!}$, which indicate the number of possible permutation scenarios if you select $r$ elements out of a whole of $n$ elements.

Answer 2

A possible difference in interpretation leads to a different result: The problem might be easier if we assume that pages could have more than one typo; in particular, that the location of each typo is independent of the location of all other typos. Then the probability of a given typo occurring within the selection of $50$ pages is $\frac{1}{10}$, and the two probabilities can be obtained using the binomial distribution: (a) would be

$$ \binom{10}{0} \left(\frac{1}{10}\right)^0 \left(\frac{9}{10}\right)^{10} = \left(\frac{9}{10}\right)^{10} $$

while (b) could be obtained most simply by noting that the probability of exactly one typo occurring within the $50$ pages would be

$$ \binom{10}{1} \left(\frac{1}{10}\right)^1 \left(\frac{9}{10}\right)^9 = \left(\frac{9}{10}\right)^9 $$

so the answer in this case would be

$$ 1 - \left(\frac{9}{10}\right)^{10} - \left(\frac{9}{10}\right)^9 $$

But note carefully that these probabilities are valid only if we assume that the locations of the typos are independent of each other. If we assume that ten pages are chosen, at random, to each contain exactly one typo, then Yujie Zha's formulation should be used.