Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Binomial and geometric distributions

  1. Dec 29, 2007 #1
    i was doing some exercises nut i'm not sure if my answers are correct
    1) X~B(5,0.25) i have to find:
    a) E(x^2) and my answer was 2.5, is this correct?
    b) P(x(>or=to)4) and my answer was 0.0889, is this correct?

    2) X~Geom(1/3) i have to find:
    a) E(x) my answer is 1/3
    b) E(x^2)
    c) var(x)
    d) P(X=4) my answer is 0.988
    e) P(X>2) my answer is..... -(1/18)
    i dont know how to compute b) and c)
    and i think d) is wrong
    and i know e) is wrong.
    can someone help please.
    Thank you very much
     
  2. jcsd
  3. Dec 29, 2007 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Would "yes" or "no" answers help you? Show what you did to get those answers and we can point out exactly where, if at all, you went wrong.
     
  4. Dec 30, 2007 #3
    my working out is:
    1) a) E(x^2)= var(x)+E(x)^2 = np(1-p+np)= 5(1/4)(1-1/4-5/4)=2.5
    b) P(x(>or = to)4)= 1-P(x=3)-P(x=2)-P(x=1)-P(x=0)
    using the formula P(x=m)=(nCm)P^m(1-p)^(n-m)
    =1-(15/1024)-(135/512)-(405/1024)-(243/1024)=91/1024=0.0889

    2) a)E(x)=1/p=1/(1/3)=3
    b) and c) i have no idea
    d) using the formula P(x=m)=P(1-P)^(m-1)
    P(x=4)=1/3(1-1/3)^3=8/81=0.988
    e) P(x>2)=1-P(x(<or=to)2)=1-P(x=2)-P(x=1)-P(x=0)
    =1-(1/3)(2/3)-(1/3)(2/3)^0-(1/3)(2/3)^-1=-1/18
    thank u very much
     
  5. Jan 2, 2008 #4

    EnumaElish

    User Avatar
    Science Advisor
    Homework Helper

    1a is correct except you meant +5/4) not -5/4).

    For 1b can you write out the nCm terms?

    2a is correct.

    For 2b use the same formula as in 1a.

    2c: Var(Geometric(p)) = (1-p)/p^2; see http://en.wikipedia.org/wiki/Geometric_distribution

    2d: 1/3(1-1/3)^3 = 8/81 < 0.1 so it cannot be > 0.9 (i.e., you have made a fraction computation error).

    2e: if the exponent is k-1 then k = 1, 2, .... OTOH, if the exponent is k, then k = 0, 1, 2, ... In either case, exponent term > 0 so it cannot be -1. See http://en.wikipedia.org/wiki/Geometric_distribution
     
  6. Jan 9, 2008 #5
    for 1b:
    my working including the ncm terms:
    P(x(>or = to)4)= 1-P(x=3)-P(x=2)-P(x=1)-P(x=0)
    using the formula P(x=m)=(nCm)P^m(1-p)^(n-m)
    =1-[(5C3)(1/4)^3(1-1/4)^2]-[(5C2)(1/4)^2(1-1/4)^3]-[(5C1)(1/4)(1-1/4)^4]-[(5C0)(1/4)^0(1/1/4)^5]
    =1-(15/1024)-(135/512)-(405/1024)-(243/1024)
    =91/1024=0.0889

    for 2e:
    i looked at the link but i still havent figured out what i had done wrong with my working out.

    thank u very much for everything else.
     
  7. Jan 9, 2008 #6

    EnumaElish

    User Avatar
    Science Advisor
    Homework Helper

    You have the exponential term -1, which shouldn't be there under either definition of the geometric distribution (referenced on the page I linked to).

    My guess is you should have 1 - (1/3)(2/3)^2 - (1/3)(2/3)^1 - (1/3)(2/3)^0, but you should verify that.

    For 1b, you have:
    1 - Binomial[5, 3] (1/4)^3 (1 - 1/4)^2 - Binomial[5, 2] (1/4)^2 (1 - 1/4)^3 - Binomial[5, 1] (1/4) (1 - 1/4)^4 - Binomial[5, 0] (1/4)^0 (1/1/4)^5, where Binomial[n,m] stands for nCm.

    Binomial[5, 3] = Binomial[5, 2] = 10
    Binomial[5, 1] = 5
    Binomial[5, 0] = 1

    so the expression is:
    1 - 10 (1/4)^3 (1 - 1/4)^2 - 10 (1/4)^2 (1 - 1/4)^3 - 5 (1/4) (1 - 1/4)^4 - 1 (1/4)^0 (1 - 1/4)^5
    =1 - 10 (1/4)^3 (3/4)^2 - 10 (1/4)^2 (3/4)^3 - 5 (1/4) (3/4)^4 - 1 (1/4)^0 (3/4)^5
    = 1 - 10 (9/1024) - 10 (27/1024) - 5 (81/1024) - 1 (243/1024)
    = 1- (90 + 270 + 405 + 243)/1024
    = 1 - 1008/1024
    = 1 - 504/512
    = 8/512
    = 1/64
    = 0.015625
     
    Last edited: Jan 9, 2008
  8. Jan 9, 2008 #7
    oh okay i see so the formula i used will only work if k=1,2,3... (as u said before)
    so if i used the formula P(x=m)=P(1-P)^(m-1)
    and did 1-P(x=2)-P(x=1)
    =1-(1/3)(2/3)-(1/3)(2/3)^0
    =4/9
    would that work?
     
  9. Jan 11, 2008 #8

    EnumaElish

    User Avatar
    Science Advisor
    Homework Helper

    Shouldn't x=3 be part of your formula?
     
  10. Jan 12, 2008 #9
    no it should only be for x=2 and x=1 as these are the ones strictly less than three
     
  11. Jan 14, 2008 #10

    EnumaElish

    User Avatar
    Science Advisor
    Homework Helper

    Your OP stated "P(x(>or = to)4)," doesn't that exclude x=3?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Binomial and geometric distributions
  1. Binomial distribution (Replies: 3)

  2. Binomial Distribution (Replies: 1)

  3. Binomial distribution (Replies: 3)

  4. Binomial distribution (Replies: 1)

Loading...