Results 1 to 7 of 7

Math Help - Proabability problem

  1. #1
    Newbie
    Joined
    Jan 2008
    Posts
    5

    Post Proabability problem

    Hi all, I`ve got an exam coming up in just over a week which will contain a question like this one.

    From the attached Venn diagram work out

    1) Determine P(A) and P(B)
    2) Determine P(B|A) and P(A|B)
    3) Determine P(B|A) and compare it with P(B|A)
    4) Determine P(B|A)
    5) Determine P(B|A) and compare it with P(B|A)
    6) Determine P(A|B)

    I don`t understand from question 3 onwards, can anybody please help me?

    cheers.................
    Attached Thumbnails Attached Thumbnails Proabability problem-venn_diagram.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1573
    Awards
    1
    3) P\left( {\neg B|A} \right) = \frac{{13}}{{824}}.

    4) P\left( {B|\neg A} \right) = \frac{{124}}{{176}}.

    Can you finish?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2008
    Posts
    19
    Quote Originally Posted by marciano1976 View Post
    Hi all, I`ve got an exam coming up in just over a week which will contain a question like this one.

    From the attached Venn diagram work out

    1) Determine P(A) and P(B)
    2) Determine P(B|A) and P(A|B)
    3) Determine P(B|A) and compare it with P(B|A)
    4) Determine P(B|A)
    5) Determine P(B|A) and compare it with P(B|A)
    6) Determine P(A|B)

    I don`t understand from question 3 onwards, can anybody please help me?

    cheers.................
    3) P(B|A) is the probability that B will occur given that A has occured. You know P(A). Now you just need the probability that B won't occur and A will, which is 13/Total. So you just do P(B and A)/P(A) = 13/824. Notice, this is 1-P(B|A)

    4) This is the probability that B will occur given that A hasn't occured. You know P(A)=1-P(A). So, you just need P(B and A), which is 124/Total. So it's just P(B and A)/P(A). Notice that this will be [P(B)-P(B|A)P(A)]/P(A) by the law of total probability

    You should get the idea now.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jan 2008
    Posts
    5

    parts 5 and 6

    Hi, Thanx for your replies to my problem.

    For part 5 i get

    52/76 = 0.2954

    For part 6 i get

    52/65 = 0.8

    i don`t think these are near the mark however.....Help!!!!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member DivideBy0's Avatar
    Joined
    Mar 2007
    From
    Melbourne, Australia
    Posts
    432
    This is what I got.......

    Let n(\varepsilon) denote the total number of elements in the universal set.

    Question 3

    Pr(B' | A)=\frac{Pr(B' \cap A)}{Pr(A)}

    From the diagram we see that Pr(B' \cap A)=\frac{13}{n(\varepsilon)}

    \Rightarrow Pr(B'|A)=\frac{(\frac{13}{n(\varepsilon)})}{(\frac  {811+13}{n(\varepsilon)})}=\frac{13}{824}

    Question 4

    Similar method to 3

    Question 5

    Pr(B'|A')=\frac{Pr(B' \cap A')}{Pr(A')}

    Pr(B' \cap A')=\frac{52}{n(\varepsilon)}

    = \frac{(\frac{52}{n(\varepsilon)})}{(\frac{824}{n(\  varepsilon)})}=\frac{52}{824}=\frac{13}{206}

    Question 6

    Pr(A'|B')=\frac{Pr(A' \cap B')}{Pr(B')}

    \Rightarrow Pr(A'|B')=\frac{(\frac{52}{n(\varepsilon)})}{(\fra  c{935}{n(\varepsilon)})}=\frac{52}{935}
    Last edited by DivideBy0; January 6th 2008 at 02:10 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1573
    Awards
    1
    DivideBy0,
    Did you missread #4?
    P\left( {B|\neg A} \right)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member DivideBy0's Avatar
    Joined
    Mar 2007
    From
    Melbourne, Australia
    Posts
    432
    Yeah sorry
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Conditional proabability from a tree diagram.
    Posted in the Statistics Forum
    Replies: 4
    Last Post: June 10th 2011, 01:09 PM
  2. Proabability With Test Questions
    Posted in the Statistics Forum
    Replies: 1
    Last Post: November 8th 2009, 11:05 AM
  3. Proabability distribution problem
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: May 25th 2009, 04:48 AM
  4. [SOLVED] [SOLVED] Proabability!
    Posted in the Statistics Forum
    Replies: 2
    Last Post: January 27th 2008, 11:51 PM
  5. blackjack proabability question
    Posted in the Statistics Forum
    Replies: 6
    Last Post: August 14th 2007, 02:42 PM

Search Tags


/mathhelpforum @mathhelpforum