Results 1 to 6 of 6

Math Help - Probability check

  1. #1
    Newbie
    Joined
    Jan 2011
    Posts
    21

    Probability check

    the probability that a surgery will succeed equals P and it's bigger than the probability that the surgery will not succeed.
    It's known that if 4 surgeries are being done the probability that exactly 2 of them will succeed is 27/128.
    Find P.

    here is the equation I got:

    6p^4-12P^3+6P^2-27/128

    FINAL EQUATION:

    6P^2-12P+5/101/128=0

    FINAL RESULT IS 3\4

    can anyone please help me correcting this equation ? I could get the required answer.
    i have to do it according to bernouli's principle.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7
    Quote Originally Posted by Mathematicsfan View Post
    the probability that a surgery will succeed equals P and it's bigger than the probability that the surgery will not succeed.
    It's known that if 4 surgeries are being done the probability that exactly 2 of them will succeed is 27/128.
    Find P.

    here is the equation I got:

    6p^4-12P^3+6P^2-27/128

    FINAL EQUATION:

    6P^2-12P+5/101/128=0

    FINAL RESULT IS 3\4

    can anyone please help me correcting this equation ? I could get the required answer.
    i have to do it according to bernouli's principle.
    6P^2(1-P)^2=\frac{27}{128}

    P^2(1-P)^2=\frac{9}{256}

    P(1-P)=\frac{3}{16} (The negative value is rejected because both factors on the LHS are positive.)

    P-P^2=\frac{3}{16}

    16P^2-16P+3=0

    16P^2-4P-12P+3=0

    4P(4P-1)-3(4P-1)=0

    (4P-1)(4P-3)=0

    P=\frac{1}{4} or P=\frac{3}{4}

    But P>\frac{1}{2} (Why?)

    Therefore, P=\frac{3}{4}.
    Last edited by alexmahone; January 6th 2011 at 09:25 AM. Reason: Changed p to P. (Yeah, I'm a perfectionist.)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by Mathematicsfan View Post
    the probability that a surgery will succeed equals P and it's bigger than the probability that the surgery will not succeed.
    It's known that if 4 surgeries are being done the probability that exactly 2 of them will succeed is 27/128.
    Find P.

    here is the equation I got:

    6p^4-12P^3+6P^2-27/128

    FINAL EQUATION:

    6P^2-12P+5/101/128=0

    FINAL RESULT IS 3\4

    can anyone please help me correcting this equation ? I could get the required answer.
    i have to do it according to bernouli's principle.
    Binomial Expansion:

    \displaystyle\ (p+q)^4=p^4+\binom{4}{1}p^3q+\binom{4}{2}p^2q^2+\b  inom{4}{3}pq^3+q^4

    The probability of exactly two successes is

    \displaystyle\binom{4}{2}p^2(1-p)^2=6p^2(1-p)^2=\frac{27}{128}

    \Rightarrow\displaystyle\ p^2q^2=\frac{9}{256}\Rightarrow\ pq=\frac{\sqrt{9}}{\sqrt{256}}=\frac{3}{16}

    pq=p(1-p)=p-p^2\Rightarrow\ 16p-16p^2=3\Rightarrow\ 16p^2-16p+3=0

    (4p)^2-4(4p)+3=0\Rightarrow\ (4p-1)(4p-3)=0

    p>(1-p)\Rightarrow\ 4p=3
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jan 2011
    Posts
    21
    thanks a lot you guys 1
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2011
    Posts
    21
    I'm totally confused, about something why did you guy use root on 9\256
    ? can you explain it to me cause I didn't study to use it, and hence I wasn't able to solve this.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by Mathematicsfan View Post
    I'm totally confused, about something why did you guy use root on 9\256
    ? can you explain it to me cause I didn't study to use it, and hence I wasn't able to solve this.
    \displaystyle\ p^2q^2=(pq)^2=\frac{9}{256}

    so we need the square root of that fraction for pq

    \displaystyle\ pq=\sqrt{(pq)^2}=\sqrt{\frac{9}{256}}=\frac{\sqrt{  9}}{\sqrt{256}}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Probability check
    Posted in the Statistics Forum
    Replies: 4
    Last Post: January 8th 2011, 05:47 AM
  2. Probability check
    Posted in the Statistics Forum
    Replies: 8
    Last Post: January 5th 2011, 05:13 AM
  3. Probability check
    Posted in the Statistics Forum
    Replies: 2
    Last Post: January 3rd 2011, 12:33 PM
  4. Probability Check
    Posted in the Statistics Forum
    Replies: 1
    Last Post: February 5th 2008, 06:31 PM
  5. check probability answer please
    Posted in the Statistics Forum
    Replies: 2
    Last Post: August 7th 2006, 08:14 PM

Search Tags


/mathhelpforum @mathhelpforum