Results 1 to 5 of 5

Math Help - Probability

  1. #1
    Newbie
    Joined
    Oct 2006
    Posts
    1

    Probability

    A bag contains the numbers 1 through 20 inclusive. If three are selected at random and placed in order from smallest to largest, find the probability that the arrangement is of three consecutive integers.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by punmaster View Post
    A bag contains the numbers 1 through 20 inclusive. If three are selected at random and placed in order from smallest to largest, find the probability that the arrangement is of three consecutive integers.
    Hello,

    I made a diagram to analyze the described situation. Maybe you can use it for your calculations

    EB
    Attached Thumbnails Attached Thumbnails Probability-numreihenfolge.gif  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by punmaster View Post
    A bag contains the numbers 1 through 20 inclusive. If three are selected at random and placed in order from smallest to largest, find the probability that the arrangement is of three consecutive integers.
    Here is another way.
    ---
    Probability is favorable to possible.

    What is favorable?
    {1,2,3} All premutations of this set = 6
    {2,3,4} All premutations of this set = 6
    .....
    {18,19,20} All premutations of this set = 6
    In total we have,
    18 Sets
    Thus,
    18\cdot 6=108
    The possible outcomes is the number of ways of selecting 3 out of twenty.
    Which is,
    _{20}C_3=1140

    Thus,
    \frac{108}{1140}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,910
    Thanks
    1759
    Awards
    1
    Quote Originally Posted by ThePerfectHacker View Post
    18 Sets Thus,
    18\cdot 6=108
    The possible outcomes is the number of ways of selecting 3 out of twenty.
    Which is,
    _{20}C_3=1140
    Thus,
    \frac{108}{1140}
    There are 18 subsets of three consecutive integers.
    There are _{20}C_3=1140 total subsets three integers.
    Thus, \frac{18}{1140} = 0.01579

    The order does not come to this.
    If we do use order then we would have \frac{108}{(20)(19)(18)} = 0.01579
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by Plato View Post
    The order does not come to this.
    You are right. When I was doing this problem I first did it with order then realized that order is not important. So I re-edited the problem but forgot to remove the 3!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: October 10th 2014, 10:02 PM
  2. Replies: 10
    Last Post: January 21st 2011, 12:47 PM
  3. Replies: 3
    Last Post: May 29th 2010, 08:29 AM
  4. Replies: 1
    Last Post: February 18th 2010, 02:54 AM
  5. Replies: 3
    Last Post: December 15th 2009, 07:30 AM

Search Tags


/mathhelpforum @mathhelpforum