Results 1 to 3 of 3

Math Help - Probability/Two fair dice are tossed

  1. #1
    Junior Member
    Joined
    Apr 2009
    Posts
    33

    Probability/Two fair dice are tossed

    Two fair dice are tossed and the product of the outcomes is recorded:

    (a)Write down the sample space s, the mode and median of S

    S={1,2,3,4,5,6,2,4,6,8,10,12,3,6,9,12,15,18,4,8,12 ,16,20,24,5,10,15,20,25,30,6,12,18,24,30,36}

    Mode: 6 and 12 (Bimodal)

    Medain=10

    (b)compute the probabilities of the events below:

    (i)Product=9

    P(9)=P(3*3)=1/36

    (ii)Product=13

    P(13)=0


    Can I write P(13)=Ф


    (iii)Product is a perfect square greater than 2

    which of my answers is true :

    P(perfect square >2)=P{(2,2) ,(3,3),(4,4)(5,5),(6,6)}=5/36

    or

    p(perfect square >2)=p(4)+p(9)+P(16)+P(25)

    =3/36 +1/36 +1/36 +1/36 +1/36=7/36

    (iv)Product=12

    p(12)=p{(3,4),(4,3),(2,6),(6,2)}=4/36

    (c)Compute the expected value for the product

    What is the meaning of expected value?and How can I obtain it?

    (d)Compute the probabilities of obtaining a double six and the expected number of attempts till a double six is attained

    Does the question mean Probability of obtaining a double 6, (6,6) is 1/36.?


    and how can i obtain the expected number of attempts till a double 6 is attained
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,234
    Thanks
    27
    Quote Originally Posted by change_for_better View Post
    Two fair dice are tossed and the product of the outcomes is recorded:

    (a)Write down the sample space s, the mode and median of S

    S={1,2,3,4,5,6,2,4,6,8,10,12,3,6,9,12,15,18,4,8,12 ,16,20,24,5,10,15,20,25,30,6,12,18,24,30,36}

    Mode: 6 and 12 (Bimodal)

    Medain=10

    (b)compute the probabilities of the events below:

    (i)Product=9

    P(9)=P(3*3)=1/36

    (ii)Product=13

    P(13)=0


    I agree with all of these

    Quote Originally Posted by change_for_better View Post





    (iii)Product is a perfect square greater than 2

    which of my answers is true :

    p(perfect square >2)=p(4)+p(9)+P(16)+P(25)

    =3/36 +1/36 +1/36 +1/36 +1/36=7/36

    (iv)Product=12

    p(12)=p{(3,4),(4,3),(2,6),(6,2)}=4/36
    These answers I also agree with



    Quote Originally Posted by change_for_better View Post







    (c)Compute the expected value for the product

    (d)Compute the probabilities of obtaining a double six and the expected number of attempts till a double six is attained
    c)

    E(X) = 1\times p(1)+2\times p(2)+ \cdots + 36 \times p(36)

    E(X) = 1\times \frac{1}{36}+2\times +\frac{2}{36} \cdots + 36 \times \frac{1}{36}


    d) \mu = n\times p where \mu = E(X) from the previous problem and p = \frac{1}{36}

    so n = \frac{\mu}{p} = \frac{E(X)}{p}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Apr 2009
    Posts
    33
    Quote Originally Posted by pickslides View Post
    I agree with all of these



    These answers I also agree with





    c)

    E(X) = 1\times p(1)+2\times p(2)+ \cdots + 36 \times p(36)

    E(X) = 1\times \frac{1}{36}+2\times +\frac{2}{36} \cdots + 36 \times \frac{1}{36}


    d) \mu = n\times p where \mu = E(X) from the previous problem and p = \frac{1}{36}

    so n = \frac{\mu}{p} = \frac{E(X)}{p}

    Thank you very much ..

    Can you explain the last point:

    d) \mu = n\times p where \mu = E(X) from the previous problem and p = \frac{1}{36}

    so n = \frac{\mu}{p} = \frac{E(X)}{p}[/quote]
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Two fair dice are thrown. Prob of event A.
    Posted in the Statistics Forum
    Replies: 5
    Last Post: March 16th 2011, 10:34 AM
  2. Roll n fair dice problem - please help
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: October 27th 2010, 04:47 AM
  3. A fair coin is tossed 10 times
    Posted in the Statistics Forum
    Replies: 2
    Last Post: April 6th 2010, 01:58 AM
  4. Replies: 1
    Last Post: March 8th 2010, 11:32 PM
  5. Replies: 2
    Last Post: December 17th 2008, 04:22 AM

Search Tags


/mathhelpforum @mathhelpforum