Results 1 to 4 of 4

Math Help - conditional probabiliity

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    15

    conditional probabiliity

    Hi, can someone check if i did this right?

    1) Two dice are rolled. We only observe that there is at least on "six" showing. What is the probability that the sum of two dice is less than 12?

    10/12

    2) find the probability that a family has two boys given that the family has at least one boy.
     \frac{\frac{1}{4} }{\frac{3}{4}} = \frac{1}{5}

    3) two dice are rolled. if the two faces are different what is the probability that at least one is six?

     \frac{11/36}{30/36} = \frac{11}{30}

    thank you, coz my stats and maths skills arent so good
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Sep 2009
    Posts
    502
    Quote Originally Posted by Pengu View Post
    Hi, can someone check if i did this right?

    1) Two dice are rolled. We only observe that there is at least on "six" showing. What is the probability that the sum of two dice is less than 12?

    10/12

    2) find the probability that a family has two boys given that the family has at least one boy.
     \frac{\frac{1}{4} }{\frac{3}{4}} = \frac{1}{5}

    3) two dice are rolled. if the two faces are different what is the probability that at least one is six?

     \frac{11/36}{30/36} = \frac{11}{30}

    thank you, coz my stats and maths skills arent so good
    Your first answer is correct.
    Your second question is incomplete. It seems that there are 5 children in the family.
    Your third answer is incorrect because you included a two sixes.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2009
    Posts
    15
    Im unsure how to do the second one, can you please help coz i thought that i did it correctly. this is how i did it though:
    S = {(b,b),(b,g),(g,b),(g,g)}.

     P(X \cap Y) = 1/4
     P(Y) = 3/4

    so im taking it that that's wrong, so how am i supposed to answer it?

    and can you please advise how im supposed to do the third one?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Sep 2009
    Posts
    502
    Quote Originally Posted by Pengu View Post
    Im unsure how to do the second one, can you please help coz i thought that i did it correctly. this is how i did it though:
    S = {(b,b),(b,g),(g,b),(g,g)}.

     P(X \cap Y) = 1/4
     P(Y) = 3/4

    so im taking it that that's wrong, so how am i supposed to answer it?

    and can you please advise how im supposed to do the third one?
    Earlier, I said your question was incomplete for the reason that we are not told how many children there are in the family. Your solution is much depended on this piece of information.

    If you had stated your problem verbatim, we must approach the problem as follows:

    Suppose that there are two children in the family, given that (at least) one of them is a boy.

    Let B = {(b,b), (b,g), (g,b)}, G = {(g,g), (b,g), (g,b)}, E=B\G={(b,b)}and S = {(b,b), (b,g), (g,b), (g,g)}

    Then





    Note: (1) If the number of children is n, your sample space, S, will contain elements.

    (2) If you know at least the younger or older is a boy, your solution will completely be different. Be sure to notice the distinction.

    (3) In you last post: If X = {(b,b)} and Y={(b,g), (g,b), (b,b)}, then your set up was partly correct. You should study my notations. For B\G, we say "Boys only". B\G = B-G.

    Your set up should be as follows:

     P(X \cap S) = 1/4
     P(Y\cap S) = 3/4
     P(X|Y) = \frac{P(X \cap S)}{P(Y\cap S)}= \frac{P(X)}{P(Y)}=1/3


    Solution #3:


    For the solution, we have two ways of doing it-- by permutation or combination.

    By permutation, we know it's a 6x6 = 36 total possible outcomes.
    Given that no faces are of the same number, we can subtract 6 out of 36, which leave us 30 total possible outcomes.

    Now we are given a condition that tells us that one of them is a six, which means that either the first or second dice is a 6:

    If the first is 6, the second cannot be 6. That leaves us 5 possible outcomes.
    If the second is 6, the first cannot be 6. That also leaves us 5 possible outcomes.

    If the 1st or 2nd is a six provide we meet all conditions, we have 10 possible outcomes.

    The probability of one of them being a 6 = \frac{10}{30}=\frac{1}{3}

    By combination:

    In this case order does not matter so there are = 30/2 = 15 total possible outcomes.

    If one is a 6, again order does not matter, so the possible outcomes with one of them being a 6 is \left(\begin{array}{c}5\\1\end{array}\right)=5

    The probability of one of them being a 6 is = \frac{5}{15}=\frac{1}{3}
    Last edited by novice; October 2nd 2009 at 08:55 PM. Reason: add solution to problem 3
    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. E(XY) and Conditional p.d.f
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: November 23rd 2010, 08:29 PM
  3. conditional
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: March 27th 2010, 02:24 PM
  4. conditional cdf
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: January 28th 2010, 05:32 PM
  5. conditional expectation
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: January 24th 2010, 12:31 AM

Search Tags


/mathhelpforum @mathhelpforum