Results 1 to 5 of 5
Like Tree1Thanks
  • 1 Post By SlipEternal

Math Help - probability of conditional events

  1. #1
    Member
    Joined
    Apr 2014
    From
    canada
    Posts
    140

    probability of conditional events

    can someone show me how to do second part and third part please?
    i really have no idea how to start it. probability of conditional events-img_20140716_055000-1-.jpg
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Feb 2014
    From
    United States
    Posts
    470
    Thanks
    192

    Re: probability of conditional events

    Quote Originally Posted by delso View Post
    can someone show me how to do second part and third part please?
    i really have no idea how to start it. Click image for larger version. 

Name:	IMG_20140716_055000[1].jpg 
Views:	7 
Size:	560.2 KB 
ID:	31299
    There are two problems with your question

    First, part 2 and part 3 are invisible in your image. In fact, not even all your answers to part 1 are visible.

    Second, even for part 1 you show no work, just answers. It is impossible for us to know what you are thinking when you just show answers.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,838
    Thanks
    707

    Re: probability of conditional events

    Write out all triples that sum to 5:

    (1,1,3), (1,3,1), (3,1,1)
    (1,2,2), (2,1,2), (2,2,1)

    There are no other triples that sum to 5. How many triples are possible? There are six possible rolls for each die. So, there are 6^3 possible triples. Hence, P(T) = \dfrac{6}{6^3} = \dfrac{1}{36}. Then, P(J|T) is the number of outcomes (among the six triples whose sum is five) that have a one in the third position. I count three. So, P(J|T) = \dfrac{3}{6} = \dfrac{1}{2}.
    Thanks from delso
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Apr 2014
    From
    canada
    Posts
    140

    Re: probability of conditional events

    sorry. i mean i manaed to get the ans for P(G2 n R3) only, but not for the rest. for P(G2 n R3) , as there're is 6 possible outcome for event G2 , but we need only 1 so 1/6 , for event R3 , there're is 6 possible outcome for it. so 1/6 x 1/6 = 1/36.... can you show me how to get P(J|R3)and P(R3|J) please? thanks..
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Apr 2014
    From
    canada
    Posts
    140

    Re: probability of conditional events

    thanks. i am able to solve the previous part using the same reasoning as in your post.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Conditional probabilty; dependent and independent events
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: October 6th 2010, 09:10 AM
  2. combined events and conditional probability
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: April 24th 2010, 01:18 AM
  3. Conditional dependency of events
    Posted in the Statistics Forum
    Replies: 0
    Last Post: January 24th 2010, 05:17 PM
  4. Conditional Probability/Sequence of Events
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: January 22nd 2009, 06:01 AM
  5. Replies: 1
    Last Post: July 9th 2008, 11:38 PM

Search Tags


/mathhelpforum @mathhelpforum