Results 1 to 11 of 11

Math Help - Prove equations.

  1. #1
    Senior Member
    Joined
    Feb 2008
    Posts
    383

    Prove equations.

    Given:

    the probability a coin lands on heads 3 times out of four tosses = 4p^3(1-p)

    the probability a coin lands on heads four times out of six tosses = 15p^4(1-p)^2
    the probability a coin lands on heads 3 times out of four tosses is twice the probability a coin lands on heads four times out of six tosses. Prove p must satisfy the equation 15p^2-15p+2=0
    thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    30p^4(1-p)^2=4p^3(1-p)

    30p^4(1-p)=4p^3

    30p^4-30p^5-4p^3=0

    30p^5-30p^4+4p^3=0

    divde by p^3
    Last edited by Archie Meade; January 14th 2010 at 12:36 PM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,993
    Thanks
    1656
    All of these follow from the basic definition of the binomial probability distribution. Do you know what that is?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by BabyMilo View Post
    Given:








    thanks.
    first, do not put questions in quotes. notice that they do not show up when i quote you? this makes it annoying.

    anyway, the trick here is to realize that you have a binomial distribution here.

    Lets say you are repeating an experiment where each outcome is independent of the others. moreover, each outcome can be considered as a "success" or "failure". Then the probability of k successes in n trials, is given by:

    P(k) = {n \choose k} p^kq^{n - k}

    where p is the probability of success and q = 1 - p is the probability of failure, and {n \choose k} = _nC_k.


    Now, for your first problem: n = 4, ~k=3 and for the second, n = 6,~k = 4. plug it in the formula i gave you to derive the desired expressions.

    for the last problem, set the expression in the first problem equal to 2 times the expression in the second and simplify to get to what you want
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Feb 2008
    Posts
    383
    Quote Originally Posted by Jhevon View Post
    according to the problem, the left side of your first equations should be multiplied by 2.
    can i just ask why left hand side and not right hand side?

    thanks
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Feb 2008
    Posts
    383
    Quote Originally Posted by Jhevon View Post
    first, do not put questions in quotes. notice that they do not show up when i quote you? this makes it annoying.

    anyway, the trick here is to realize that you have a binomial distribution here.

    Lets say you are repeating an experiment where each outcome is independent of the others. moreover, each outcome can be considered as a "success" or "failure". Then the probability of k successes in n trials, is given by:

    P(k) = {n \choose k} p^kq^{n - k}

    where p is the probability of success and q = 1 - p is the probability of failure, and {n \choose k} = _nC_k.


    Now, for your first problem: n = 4, ~k=3 and for the second, n = 6,~k = 4. plug it in the formula i gave you to derive the desired expressions.

    for the last problem, set the expression in the first problem equal to 2 times the expression in the second and simplify to get to what you want
    sry i thought it would be easier to read.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Feb 2008
    Posts
    383
    Quote Originally Posted by HallsofIvy View Post
    All of these follow from the basic definition of the binomial probability distribution. Do you know what that is?
    yes this is of a Statistics module. thanks!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Yes,

    in the question, it's stated,
    the probability of the 3 out of 4 case is twice the probabilty of the 4 out of 6 case.

    Therefore

    4p^3(1-p)

    is twice the other.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by BabyMilo View Post
    can i just ask why left hand side and not right hand side?

    thanks
    did you read the problem? it told us one was twice the other. there is a specific one that should be multiplied by 2.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Senior Member
    Joined
    Feb 2008
    Posts
    383
    Quote Originally Posted by Archie Meade View Post
    Yes,

    in the question, it's stated,
    the probability of the 3 out of 4 case is twice the probabilty of the 4 out of 6 case.

    Therefore

    4p^3(1-p)

    is twice the other.
    then shouldnt it be 15p^4(1-p)^2=2(4p^3(1-p))
    Follow Math Help Forum on Facebook and Google+

  11. #11
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by BabyMilo View Post
    then shouldnt it be 15p^4(1-p)^2=2(4p^3(1-p))
    if you say "A is twice B" you are saying "A = 2B"

    now look at what your A and B are.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Using substitution to prove two equations
    Posted in the Calculus Forum
    Replies: 7
    Last Post: March 14th 2011, 06:39 AM
  2. Using Homogeneous Equations to prove the problems ?
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: July 25th 2010, 05:12 AM
  3. Replies: 5
    Last Post: April 4th 2010, 04:53 PM
  4. Replies: 1
    Last Post: April 30th 2009, 01:46 PM
  5. stiff differential equations [prove]
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 31st 2008, 08:58 AM

Search Tags


/mathhelpforum @mathhelpforum