Results 1 to 5 of 5

Math Help - 123 written in a random order probability any are in the right place

  1. #1
    Junior Member
    Joined
    Oct 2009
    From
    Aberystwyth, Wales
    Posts
    36

    123 written in a random order probability any are in the right place

    i have the question if 1,2 & 3 are written in a random order whats the chance one or more of the didgits are in the right place

    i know the probability is 2/3 as i can easily work it out in my head but i have no idea how to work it out mathematically ive tryed loads of different ways but alway end up with something greater than 1
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Nov 2009
    Posts
    2
    There are 3! possible ways to arrange the numbers.

    There are (2)2! ways that at least one letter will be in the correct position.

    (when one number is held fixed there are 2! ways to arrange the other numbers)

    2!2!/3! = 4/6 = 2/3
    Last edited by Kivit; November 16th 2009 at 03:07 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,685
    Thanks
    616
    Hello, renlok!

    Kivit got the right answer, but I don't understand his reasoning.


    If 1,2 & 3 are written in a random order
    what's the chance one or more of the digits are in the right place?
    There are 3! = 6 possible permutations.


    There are 3 ways that exactly one digit is in the right place: .132, 321, 213

    There is no way that exactly two digits are in the right place.

    There is 1 way that all three digits are in the right place: 123

    Hence, there are 4 ways that at least one digit is in the right place.


    The probability is: . \frac{4}{6} \:=\:\frac{2}{3}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Here's another approach . . . a rather primitive one.

    There are 3! = 6 possible permutations.

    How many of them have no digits in the right place?


    The first digit cannot be "1"; it must be "2" or "3".

    If the first digit is "2": . 2 _ _
    . . we must place the "1" and "3".
    The "3" cannot be the third digit; it must be the second digit.
    . . We have: .231

    If the first digit is "3": .3 _ _
    . . we must place the "1" and "2".
    The "2" cannot be the second digit; it must be the third digit.
    . . We have: .312

    There are only two ways that no digits are in the right place.

    Hence, there are: . 6-2 \:=\:4 ways with at least one digit correct.

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Oct 2009
    From
    Aberystwyth, Wales
    Posts
    36
    is there no way to do it with nPr and nCr becausee the next question is the same but with n numbers
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Nov 2009
    Posts
    2
    (n-1)(n-1)!/n!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Calculating Nth place from 1st place probability
    Posted in the Advanced Statistics Forum
    Replies: 10
    Last Post: February 7th 2011, 04:24 PM
  2. Probability and random random varible question
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: May 15th 2010, 08:53 PM
  3. Replies: 1
    Last Post: May 2nd 2010, 01:10 PM
  4. CD random order problem
    Posted in the Statistics Forum
    Replies: 9
    Last Post: March 31st 2009, 02:39 AM
  5. Order Statistics, N independent uniform random variables
    Posted in the Advanced Statistics Forum
    Replies: 9
    Last Post: March 22nd 2009, 10:12 PM

Search Tags


/mathhelpforum @mathhelpforum