Thread: prob that at least one was correct

1. prob that at least one was correct

Suppose that an absentmined secretary wrote n letters and sealed them in envelops before writing the addresses on the envelops. Then she wrote the n addresses on the envelops at random. What is the prob that at least one letter was addressed correctly?

Im thinking to find the prob that no one was addressed right then subtract it from 1, and the chance address the first one right is 1/5, and the second one is 1/4, what if the first one was wrong already?

2. Originally Posted by wopashui
Suppose that an absentmined secretary wrote n letters and sealed them in envelops before writing the addresses on the envelops. Then she wrote the n addresses on the envelops at random. What is the prob that at least one letter was addressed correctly?

Im thinking to find the prob that no one was addressed right then subtract it from 1, and the chance address the first one right is 1/5, and the second one is 1/4, what if the first one was wrong already?
You are right. P(at least one was addressed correctly) = 1 - P(none were addressed correctly)

3. Originally Posted by wopashui
Suppose that an absentmined secretary wrote n letters and sealed them in envelops before writing the addresses on the envelops. Then she wrote the n addresses on the envelops at random. What is the prob that at least one letter was addressed correctly? Im thinking to find the prob that no one was addressed right then subtract it from 1, and the chance address the first one right is 1/5, and the second one is 1/4, what if the first one was wrong already?
You idea is correct but answer is not.
The is a derangement problem. The derangement of five is 44 ways.

4. sry, it's not 1/5, it's 1/n, so the prob of getting none of them right is ($\displaystyle \dfrac {n-1}{n}$) ($\displaystyle \dfrac {n-2}{n-1}$).....($\displaystyle \dfrac {1}{2})$ then subtract this from 1, is that correct?

5. The correct answer is $\displaystyle 1-\frac{44}{120}$.

6. Originally Posted by Plato
The correct answer is $\displaystyle 1-\frac{44}{120}$.

what if there are n letters and n address, not 5

7. Originally Posted by wopashui
what if there are n letters and n address, not 5
A derangement is permutation in which every element is active (none remains fixed).

Here is the formula for the number of derangements on n objects.
$\displaystyle D(n) = n! \cdot \sum\limits_{k = 0}^n {\frac{{\left( { - 1} \right)^k }}{{k!}}} \approx \frac{{n!}}{e}$

Therefore to answer your question, a good approximation is $\displaystyle 1-\dfrac{n!}{e}$.

8. Originally Posted by Plato
A derangement is permutation in which every element is active (none remains fixed).

Here is the formula for the number of derangements on n objects.
$\displaystyle D(n) = n! \cdot \sum\limits_{k = 0}^n {\frac{{\left( { - 1} \right)^k }}{{k!}}} \approx \frac{{n!}}{e}$

Therefore to answer your question, a good approximation is $\displaystyle 1-\dfrac{n!}{e}$.
this makes sense to me, however we have not cover any material of derangement yet, is there any other way of solving this, why is (n-1)!/n! not right?