Results 1 to 8 of 8

Math Help - prob that at least one was correct

  1. #1
    Senior Member
    Joined
    Jan 2010
    Posts
    273

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member MacstersUndead's Avatar
    Joined
    Jan 2009
    Posts
    291
    Thanks
    32
    Quote Originally Posted by wopashui View Post
    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)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,956
    Thanks
    1780
    Awards
    1
    Quote Originally Posted by wopashui View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Jan 2010
    Posts
    273
    sry, it's not 1/5, it's 1/n, so the prob of getting none of them right is ( \dfrac {n-1}{n}) ( \dfrac {n-2}{n-1}).....( \dfrac {1}{2}) then subtract this from 1, is that correct?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,956
    Thanks
    1780
    Awards
    1
    The correct answer is 1-\frac{44}{120}.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Jan 2010
    Posts
    273
    Quote Originally Posted by Plato View Post
    The correct answer is 1-\frac{44}{120}.

    what if there are n letters and n address, not 5
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,956
    Thanks
    1780
    Awards
    1
    Quote Originally Posted by wopashui View Post
    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.
    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 1-\dfrac{n!}{e}.
    Last edited by Plato; October 1st 2010 at 04:05 PM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Jan 2010
    Posts
    273
    Quote Originally Posted by Plato View Post
    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.
    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 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?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 15
    Last Post: July 29th 2011, 02:39 AM
  2. Replies: 2
    Last Post: April 12th 2011, 11:13 AM
  3. Replies: 1
    Last Post: October 8th 2009, 11:53 PM
  4. UDD prob
    Posted in the Business Math Forum
    Replies: 1
    Last Post: May 10th 2009, 11:25 AM
  5. Log Prob.
    Posted in the Algebra Forum
    Replies: 3
    Last Post: February 3rd 2009, 05:14 PM

Search Tags


/mathhelpforum @mathhelpforum