Results 1 to 6 of 6
Like Tree1Thanks
  • 1 Post By Plato

Math Help - Basic Statistic work

  1. #1
    Banned
    Joined
    Oct 2009
    Posts
    39
    Thanks
    1

    Basic Statistic work

    A, B, C are events of random experiment, so that:

    P(A)=0.45
    P(B)=0.4
    P(C)=0.3
    P(A \cap B \cap C)=0.07
    P(A' \cap B \cap C)=0.08
    P(A \cap B' \cap C)=0.1
    P(A \cap B \cap C')=0.03

    Find the following;

    P(A-B)=?,
    P(A \cap B' \cap C')=?
    P(A' \cup B' \cup C')=?
    P(A \cup B')=?
    P(A \cup B \cup C)=?
    P(A' \cap B' \cap C')=?

    So, here are my answers.... Could anyone please check them ? It really took me one whole night to solve.
    Please bear with me... for not using latex on my answers as well, but it really takes a lot time, since i am new to latex.

    Page one = Imageshack - 58168563.png
    Page two = Imageshack - 85307415.png



    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    39
    Thanks
    1

    Re: Basic Statistic work

    So, i decided to bump this thread up. I am back without scanned papers. Just Latex. For the obvious reasons ill just provide only 2 solutions.

    P(A-B)=?

    I solved it this way:

    P(A-B)= P(A) - P(A \cap B) = P(A) - [ P(A \cap B \cap C) + P(A \cap B \cap C')] =...= 0.35


    P(A \cap B' \cap C')=?

    ans:
    Using DeMorgan we get to this point:
    P(A \cap B' \cap C')= 1-P(A' \cup B \cup C)=

    = 1-[P(A')+P(B)+P(C)-P(A' \cap B)-P(A' \cap C)-P(B \cap C)+P(A' \cap B \cap C)]

    = 1-[P(A')+P(B)+P(C)-(P(B)-P(A \cap B))-(P(C)-P(A \cap C))-P(B \cap C)+P(A' \cap B \cap C)]

    After simplication, i am getting here:

    = 1-[1-P(A)+P(A \cap B)+P(A \cap C)-P(B \cap C)+P(A' \cap B \cap C)]
    But:

    P(A \cap B)=P(A \cap B \cap C)+P(A \cap B \cap C')=0.1
    P(A \cap C)=P(A \cap B \cap C)+P(A \cap B' \cap C)=0.17
    P(B \cap C)=P(A \cap B \cap C)+P(A' \cap B \cap C)=0.15

    We know the rest values from given so i conclude =0.25


    My Big Question, given all these facts , is my analysis correct ? The question says that A,B,C are events of a random experiment therefore i think that A,B,C ⊆ Ω , so my analysis should be correct. Any1 to confirm ?

    B)

    And something else, i am learning combinations as well. So lets say we have this standard example: we have this word --> 'ABCADEFGHEI' 11 Words in total. 2*A and 2*E . How many anagrams of this word can we do provided that the 1st letter is always G and both E's will be next each other.

    So, without much thought i came up to this: 9!/2! . But wouldnt an answer like (9!*2!/2!) make more sense? My point is this; the boxed EE's can be arranged in 2! ways (sounds stupid i know, the result is always the same EE or EE). So, 11 letters - 1 (Letter G) =10 Letters . Then, the EE's are boxed so it counts as a single letter, thus we getting a total of 9 letters. And i divide by 2! due to the AA's.

    What am i missing?

    Thanks a lot.
    Last edited by primeimplicant; March 16th 2012 at 03:37 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,901
    Thanks
    1756
    Awards
    1

    Re: Basic Statistic work

    Quote Originally Posted by primeimplicant View Post
    P(A-B)=?
    I solved it this way:
    P(A-B)= P(A) - P(A \cap B) = P(A) - [ P(A \cap B \cap C) + P(A \cap B \cap C')] =...= 0.35
    P(A \cap B' \cap C')=? ans:
    Using DeMorgan we get to this point:
    P(A \cap B' \cap C')= 1-P(A' \cup B \cup C)=
    = 1-[P(A')+P(B)+P(C)-P(A' \cap B)-P(A' \cap C)-P(B \cap C)+P(A' \cap B \cap C)]
    = 1-[P(A')+P(B)+P(C)-(P(B)-P(A \cap B))-(P(C)-P(A \cap C))-P(B \cap C)+P(A' \cap B \cap C)]
    After simplication, i am getting here:
    = 1-[1-P(A)+P(A \cap B)+P(A \cap C)-P(B \cap C)+P(A' \cap B \cap C)]
    But:
    P(A \cap B)=P(A \cap B \cap C)+P(A \cap B \cap C')=0.1
    P(A \cap C)=P(A \cap B \cap C)+P(A \cap B' \cap C)=0.17
    P(B \cap C)=P(A \cap B \cap C)+P(A' \cap B \cap C)=0.15
    That work is correct!

    Quote Originally Posted by primeimplicant View Post
    B) And something else, i am learning combinations as well. So lets say we have this standard example: we have this word --> 'ABCADEFGHEI' 11 Words in total. 2*A and 2*E . How many anagrams of this word can we do provided that the 1st letter is always G and both E's will be next each other.
    Please rewrite part b).
    Just state the actual question! Do not try to explain it just state the exact question.
    Thanks from mash
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Oct 2009
    Posts
    39
    Thanks
    1

    Re: Basic Statistic work

    Quote Originally Posted by Plato View Post
    Please rewrite part b).
    Just state the actual question! Do not try to explain it just state the exact question.
    given this word 'ABCADEFGHEI'

    How many different words can we produce from the letters of the above word (anagrams), if the first letter is G and letters EE are together. for example "GBACADEEFGHI" and "GBACADFGEEHI" are acceptable words, while "BGACADEFEGHI" is not acceptable.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,901
    Thanks
    1756
    Awards
    1

    Re: Basic Statistic work

    Quote Originally Posted by primeimplicant View Post
    given this word 'ABCADEFGHEI'
    How many different words can we produce from the letters of the above word (anagrams), if the first letter is G and letters EE are together. for example "GBACADEEFGHI" and "GBACADFGEEHI" are acceptable words, while "BGACADEFEGHI" is not acceptable.
    Because we know where the G must be, we disregard it. Same for the second E.
    In effect we are rearranging the string AABCDEFHI:~\frac{9!}{2!}~.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Banned
    Joined
    Oct 2009
    Posts
    39
    Thanks
    1

    Re: Basic Statistic work

    Quote Originally Posted by Plato View Post
    Because we know where the G must be, we disregard it. Same for the second E.
    In effect we are rearranging the string AABCDEFHI:~\frac{9!}{2!}~.
    Hmm i see, just like my original thought. Thanks.

    Edit: But if EE's were different colours and must be together. For example AABCDEEFHI , AABCDEEFHI, shouldnt we have ({9!}*{2!})/2! different words ?
    Last edited by primeimplicant; March 17th 2012 at 05:41 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Basic Work Problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 10th 2011, 10:27 PM
  2. Replies: 2
    Last Post: August 15th 2010, 02:12 AM
  3. Replies: 2
    Last Post: November 4th 2009, 06:03 AM
  4. another statistic
    Posted in the Statistics Forum
    Replies: 5
    Last Post: July 22nd 2009, 06:43 AM
  5. Statistic Help
    Posted in the Advanced Statistics Forum
    Replies: 5
    Last Post: January 27th 2008, 02:02 PM

Search Tags


/mathhelpforum @mathhelpforum