Results 1 to 14 of 14

Math Help - Binomial distribution

  1. #1
    Newbie
    Joined
    Jan 2011
    Posts
    7

    Binomial distribution

    Hi all.

    This is my first forum post, so please forgive me if I have done something wrong.

    Now for my problem:
    A stochastic variable X is binomial distributed with number of trials n = 18 and success probability in each trial p = 0.4.
    Calculate the probability that X is less than 11 when given the fact that X is greather than or equal to 7.

    My suggestion:
    P(7 \leq X < 11) = P(X > 11) - P(X \geq 7) = P(X \geq 11) - P(X \geq 7) \simeq 0.4163
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor harish21's Avatar
    Joined
    Feb 2010
    From
    Dirty South
    Posts
    1,036
    Thanks
    10
    X is less than 11 when given the fact that X is greater than or equal to 7 .

    use conditional distribution...

    P(X<11|X \geq 7)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2011
    Posts
    7
    Quote Originally Posted by harish21 View Post
    X is less than when given the fact that X is greather than or equal to .

    use conditional distribution...

    P(X<11|X \geq 7)
    Thank you for your time! I am really bad at probability and statistics, so I am fare from sure of what to do. I guess I have to use Bayes' formula:

    P(X < 11 \mid X \geq 7) = P(X \geq 7 \mid X < 11) \cdot \frac{P(X < 11)}{P(X \geq 7)} = P(X \geq 7 \mid X < 11) \cdot \frac{1 - P(X \geq 11)}{P(X \geq 7)}

    but how do I calculate P(X \geq 7 \mid X < 11)?

    If it isn't correct, I would very much appreciate if you would please tell me the solution.

    Thank you in advance!

    PS. Maiden rules!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,648
    Thanks
    1596
    Awards
    1
    P(X<11|X\ge 7)=\dfrac{P(7\le X <11)}{P(X\ge 7)}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2011
    Posts
    7
    Quote Originally Posted by Plato View Post
    P(X<11|X\ge 7)=\dfrac{P(7\le X <11)}{P(X\ge 7)}
    So
    P(X < 11 \mid X \geq 7) = \dfrac{P(7 \leq X < 11)}{P(X \geq 7)} = \dfrac{P(X \geq 11) - P(X \geq 7)}{P(X \geq 7)} \simeq 0.7388
    or what?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by SvendMortensen View Post
    So
    P(X < 11 \mid X \geq 7) = \dfrac{P(7 \leq X < 11)}{P(X \geq 7)} = \dfrac{P(X \geq 11) - P(X \geq 7)}{P(X \geq 7)} \simeq 0.7388
    or what?
    No. You should know that \Pr(7 \leq X < 11) is equal to \Pr(X < 11) - \Pr(X < 7).


    Edit: Corrected typo pointed out by Plato in order to avoid confusing the OP.
    Last edited by mr fantastic; January 30th 2011 at 12:05 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,648
    Thanks
    1596
    Awards
    1
    Right idea. Wrong concept.
    P(7\le X < 11)=P(X<11)-P(X<7)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Jan 2011
    Posts
    7
    Quote Originally Posted by Plato View Post
    Right idea. Wrong concept.
    P(7\le X < 11)=P(X<11)-P(X<7)
    Of course--a typo from my side; is

    P(X < 11 \mid X \geq 7) = \dfrac{P(X \leq 11) - P(X \leq 7)}{P(X \leq 7)} \simeq \dfrac{0.9797-0.5634}{0.5634} \simeq 0.7388

    correct?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,648
    Thanks
    1596
    Awards
    1
    Quote Originally Posted by SvendMortensen View Post
    Now for my problem:
    A stochastic variable X is binomial distributed with number of trials n = 18 and success probability in each trial p = 0.4.
    Calculate the probability that X is less than [tex]11[/MATh] when given the fact that X is greather than or equal to 7.
    Look at the OP. Are those wrong?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Jan 2011
    Posts
    7
    Quote Originally Posted by Plato View Post
    Look at the OP. Are those wrong?
    Yet another typo from my side; I meant

    P(X < 11 \mid X \geq 7) = \dfrac{P(7 \leq X < 11)}{P(X \geq 7)} = \dfrac{P(X \leq 11) - P(X \leq 7)}{1 - P(X \leq 7)} \simeq \dfrac{0.9797-0.5634}{1-0.5634} \simeq 0.9535
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,648
    Thanks
    1596
    Awards
    1
    You have yet another mistake.

    P(X\ge 7)=1-P(X<7)
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    Jan 2011
    Posts
    7
    Now I am (even more) confused! If my suggestion in my latest post isn't correct, I don't know that to do! Is it then

    P(X < 11 \mid X \geq 7) = \dfrac{P(7 \leq X < 11)}{P(X \geq 7)} = \dfrac{P(X < 11) - P(X \leq 7)}{1 - P(X < 7)} = \dfrac{P(X \leq 10) - P(X \leq 7)}{1 - P(X \leq 6)} \simeq \dfrac{0.9424-0.5634}{1-0.3743} \simeq 0.6056

    or am I doing yet another thing wrong?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,648
    Thanks
    1596
    Awards
    1
    P(7\le X<11)=P(X<11)-P(X<7)


    This is the correct answer: 0.9078085

    Over and out.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Newbie
    Joined
    Jan 2011
    Posts
    7
    Thanks!

    P(X < 11 \mid X \geq 7) = \dfrac{P(X \leq 10) - P(X \leq 6)}{1 - P(X \leq 6)} \simeq 0.9078
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: March 21st 2010, 05:25 PM
  2. Replies: 1
    Last Post: November 12th 2009, 12:38 AM
  3. Replies: 1
    Last Post: March 11th 2009, 11:09 PM
  4. Cumulative distribution function of binomial distribution
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: October 31st 2008, 03:34 PM
  5. Binomial distribution help!
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: July 17th 2008, 08:18 PM

Search Tags


/mathhelpforum @mathhelpforum