Results 1 to 5 of 5

Math Help - Probability

  1. #1
    Newbie
    Joined
    Jul 2009
    Posts
    10

    Probability

    p(k) denotes the probability of exactly k successes in n independent trials of a random experiment in which the probability of a success is p. q=1-p
    How do you show that :
    p(k+1) = (n-k)p , k=0,1,2,...,n-1 p(k) (k+1)q
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,617
    Thanks
    1581
    Awards
    1
    Quote Originally Posted by Pink Lady View Post
    p(k+1) = (n-k)p , k=0,1,2,...,n-1 p(k) (k+1)q
    Can you tell me what the above is.
    I really have no idea what is could possibly mean.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    p(k) = {n\choose k}p^kq^{n-k}

    so I guess you need to play with

    p(k+1) = {n\choose k+1}p^{k+1}q^{n-k-1}

    But I'm not too sure what this.... (n-k)p , k=0,1,2,...,n-1 p(k) (k+1)q.... is
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jul 2009
    Posts
    10
    It was meant to be written as follows:
    p(k+1) / p(k) = (n-k)p / (k+1)q , when k=0,1,2,3,...,n-1 and q=1-p

    Sorry about the confusion
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    Well, that follows directly from what I wrote.
    Just divide the two and use x!/(x-1)!=x twice.
    AND it's k=0,1,..,n not n-1.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 10
    Last Post: January 21st 2011, 11:47 AM
  2. Replies: 3
    Last Post: May 29th 2010, 07:29 AM
  3. fundamental in probability & convergence with probability 1
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: February 23rd 2010, 09:58 AM
  4. Replies: 1
    Last Post: February 18th 2010, 01:54 AM
  5. Replies: 3
    Last Post: December 15th 2009, 06:30 AM

Search Tags


/mathhelpforum @mathhelpforum