Results 1 to 6 of 6

Math Help - Conditional probability

  1. #1
    Junior Member
    Joined
    Aug 2009
    Posts
    26

    Conditional probability

    Rachel and annah play each other at badminton. each games results in either a win for Rachel or a win for Annah, the P of rach winning the first game is 0.6, If rach wins the first game the prob of her winning the next game is 0.7 but if she loses the P is 0.4

    Find conditional prob that rach wins the first game given that she loses the second one....

    so P (AUB) = wins 1st and loses second... = 0.6 x 0.3 = 0.18

    And prob she wins the first game = 0.6

    So the Conditional prob = 0.18/0.6 = 0.3 ?

    The answer is 0.429 ?

    Help ?

    and second part, find P that rachel wins 2 games and loses 1 game out of the first free games she plays ?

    the p of only 2 match is given ? :s all confused...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,789
    Thanks
    1684
    Awards
    1
    If I follow the question:
    You are asked to find P(R_1|R_2^{~c}).

    Now P(R_2^{~c})=P(R_1R_2^{~c})+P(R_1^{~c}R_2^{~c})=(.6  )(.3)+(.4)(.4).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2009
    Posts
    26
    The first part the answer is 0.429, and the way i learnt conditional probability is, P(A U B)/A....

    And for the second part, that leaves me some choices i think, like she wins first, then loses second and wins third,
    and etc etc, so could be a combination ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,789
    Thanks
    1684
    Awards
    1
    Quote Originally Posted by ramzel View Post
    The first part the answer is 0.429, and the way i learnt conditional probability is, P(A U B)/A....
    First of all, I disagree with that given answer.
    The answer I get is \color{blue}0.529 so I think that there is a typo in your text: the 4 should be 5.

    Next I do not think that you fully understand the notation of conditional probability.
    It involves intersection not union: P(A|B)=\frac{P(A\cap B)}{P(B)}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Aug 2009
    Posts
    26
    I got it the answer i got is 0.429,

    P she loses second game = 0.18 + 0.24

    P she wins first and loses second = 0.18

    0.18/0.42 = 0.42857

    thank you Plato for your help,

    And the second part ? how do i do it
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,789
    Thanks
    1684
    Awards
    1
    ok
    Last edited by Plato; October 8th 2009 at 05:12 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: July 22nd 2011, 01:39 AM
  2. Conditional probability?
    Posted in the Statistics Forum
    Replies: 7
    Last Post: June 17th 2011, 03:38 PM
  3. Conditional Probability using the Law of Total Probability
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: October 7th 2010, 03:01 AM
  4. Conditional Probability
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: May 16th 2010, 08:42 PM
  5. Continuous probability - conditional probability
    Posted in the Statistics Forum
    Replies: 1
    Last Post: October 1st 2009, 01:21 AM

Search Tags


/mathhelpforum @mathhelpforum