Results 1 to 5 of 5

Thread: Need help on some problems

  1. #1
    Newbie
    Joined
    May 2018
    From
    Turkey
    Posts
    5

    Question Need help on some problems

    If P(A/B)=P(A/B') then show that A and B are independent?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    6,356
    Thanks
    2732

    Re: Need help on some problems

    given $P[A|B] = P[A | B']$

    Note that clearly $B \cup B' = U$ and $B \cap B' = \emptyset$ so $P[B]+P[B']=1$

    $\begin{align*}

    &P[A] = \\ \\

    &P[A|B]P[B] + P[A|B']P[B'] = \\ \\

    &P[A|B]P[B] + P[A|B]P[B'] = \\ \\

    &P[A|B](P[B]+P[B'])= \\ \\

    &P[A|B]

    \end{align*}$

    multiply both sides by $P[B]$

    $P[A]P[B] = P[A|B]P[B] = P[A \wedge B]$

    and thus $A$ and $B$ are independent.
    Last edited by romsek; May 28th 2018 at 08:34 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    22,146
    Thanks
    3034
    Awards
    1

    Re: Need help on some problems

    Quote Originally Posted by lera View Post
    If P(A/B)=P(A/B') then show that A and B are independent?
    $\begin{gathered}
    P(A|B) = P(A|B') \hfill \\
    \frac{{P(A \cap B)}}{{P(B)}} = \frac{{P(A \cap B')}}{{P(B')}} \hfill \\
    \frac{{P(A \cap B)}}{{P(B)}} = \frac{{P(A \cap B')}}{{1 - P(B)}} \hfill \\
    [1 - P(B)]P(A \cap B) = P(B)P(A \cap B') \hfill \\
    P(A \cap B) = P(B)[P(A \cap B) + P(A \cap B')] \hfill \\
    P(A \cap B) = P(B)P(A) \hfill \\
    \hfill \\
    \end{gathered} $
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    May 2018
    From
    Turkey
    Posts
    5

    Re: Need help on some problems

    Quote Originally Posted by romsek View Post
    given $P[A|B] = P[A | B']$

    Note that clearly $B \cup B' = U$ and $B \cap B' = \emptyset$ so $P[B]+P[B']=1$

    $\begin{align*}

    &P[A] = \\ \\

    &P[A|B]P[B] + P[A|B']P[B'] = \\ \\

    &P[A|B]P[B] + P[A|B]P[B'] = \\ \\

    &P[A|B](P[B]+P[B'])= \\ \\

    &P[A|B]

    \end{align*}$

    multiply both sides by $P[B]$

    $P[A]P[B] = P[A|B]P[B] = P[A \wedge B]$

    and thus $A$ and $B$ are independent.
    Thanks !!!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    May 2018
    From
    Turkey
    Posts
    5

    Re: Need help on some problems

    Quote Originally Posted by Plato View Post
    $\begin{gathered}
    P(A|B) = P(A|B') \hfill \\
    \frac{{P(A \cap B)}}{{P(B)}} = \frac{{P(A \cap B')}}{{P(B')}} \hfill \\
    \frac{{P(A \cap B)}}{{P(B)}} = \frac{{P(A \cap B')}}{{1 - P(B)}} \hfill \\
    [1 - P(B)]P(A \cap B) = P(B)P(A \cap B') \hfill \\
    P(A \cap B) = P(B)[P(A \cap B) + P(A \cap B')] \hfill \\
    P(A \cap B) = P(B)P(A) \hfill \\
    \hfill \\
    \end{gathered} $
    Thanks !!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 9
    Last Post: Nov 1st 2016, 11:29 PM
  2. Counting Problems Math Help-3 word problems
    Posted in the Statistics Forum
    Replies: 9
    Last Post: Nov 15th 2012, 08:30 PM
  3. Replies: 0
    Last Post: Apr 2nd 2012, 05:11 PM
  4. binomial problems/normal problems
    Posted in the Statistics Forum
    Replies: 1
    Last Post: Oct 20th 2010, 12:46 AM
  5. binomial problems as normal problems
    Posted in the Statistics Forum
    Replies: 1
    Last Post: Oct 20th 2010, 12:41 AM

Search Tags


/mathhelpforum @mathhelpforum