Results 1 to 7 of 7

Math Help - Probability question please

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    10

    Probability question please

    Four dice are tossed, what is the probability of getting one pair and 2 other single number?

    I try 6*5C2*????
    That is all i got so far and I dont know how to figure out the rest.

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Jason2009 View Post
    Four dice are tossed, what is the probability of getting one pair and 2 other single number?

    I try 6*5C2*????
    That is all i got so far and I dont know how to figure out the rest.

    You need the number of ways of getting two numbers the same and the other numbers different divided by the total number of ways.

    To get the number of ways of getting two numbers the same and the other numbers different:

    First use the pigeon hole principle: 1 \times 1 \times 5 \times 4 = 20.

    Now arrange the above: \frac{4!}{2!} = 12.

    So the number of ways of getting two numbers the same and the other numbers different is equal to 240.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2009
    Posts
    10
    No. Why answer is 720???
    Really confused.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Jason2009 View Post
    No. Why answer is 720???
    Really confused.
    Please post something that makes sense. What do you mean by "No"? Where has 720 come from? State clearly what you don't understand.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Apr 2009
    Posts
    10
    The answer of this question is 720 , and I have no idea how to figure it out.
    240 is not right.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,676
    Thanks
    608
    Hello, Jason2009!

    Okay, let's baby-step through it . . .


    Four dice are tossed, what is the probability of getting one pair and 2 single numbers?

    Two of the dice will have the same number . . . which two?
    . . There are: . {4\choose2} \:=\:{\color{blue}6} choices.

    The first die can have any number: .Prob = {\color{blue}\frac{6}{6}}

    The second die must have the same number: .Prob = {\color{blue}\frac{1}{6}}

    The third die must have a different number: .Prob = {\color{blue}\frac{5}{6}}

    The fourth die must have yet another different number: .Prob = {\color{blue}\frac{4}{6}}


    Therefore, the answer is: . 6\cdot\frac{6}{6}\cdot\frac{1}{6}\cdot\frac{5}{6}\  cdot\frac{4}{6} \;=\;\frac{720}{1296} \;=\;\boxed{{\color{blue}\frac{5}{9}}}

    Follow Math Help Forum on Facebook and Google+

  7. #7
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by mr fantastic View Post
    You need the number of ways of getting two numbers the same and the other numbers different divided by the total number of ways.

    To get the number of ways of getting two numbers the same and the other numbers different:

    First use the pigeon hole principle: 1 \times 1 \times 5 \times 4 = 20.

    Now arrange the above: \frac{4!}{2!} = 12.

    So the number of ways of getting two numbers the same and the other numbers different is equal to 240.
    Well, I got careless here. Nevertheless, with a small bit of thought it should have been clear how to fix things up.

    First of all, \frac{4!}{2!} = 6, not 12. You should have spotted that carelessness straight off. Then you multiply by 6 for the 6 possible numbers. So you end up with ..... 720, not 240.

    See, all it takes is a bit of willingness to engage with and try to understand what gets posted, instead of saying "240 isn't right and I don't know what to do".
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: February 10th 2013, 02:11 PM
  2. Probability question involving (A given B type question )
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: November 9th 2009, 09:08 AM
  3. Probability Question
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: September 2nd 2009, 06:57 AM
  4. Probability Question
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: November 25th 2007, 08:16 PM
  5. A probability question and an Expectations question
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: January 29th 2006, 06:13 PM

Search Tags


/mathhelpforum @mathhelpforum