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  1. Submitted By: Al — September 18, 2007
    +3 votes
      + -

    Assume Prob(boy) = Prob(girl) = 1/2.
    Prob(1 kid) = Prob(boy) = 1/2, Prob(2 kids) = Prob(girl)*Prob(boy) = (1/2)*(1/2) = 1/4, …, Prob(N kids) = (1/2)^N.
    The proportion of boys to girls (b/g) for the country is the inverse of the expectation of the girls to boys (g/b). For 1 kid, we have 1 boy (b=1) and 0 girls (g=0) and an expected g/b of (g/b)*Prob(1 kid) = 0*1/2. For 2 kids, the expected g/b is g/b*Prob(2) = (1/1)*(1/4). For 3 kids, (2/1)*(1/8). For N kids, (N-1)*(1/2^N)
    Thus, expectation of g/b = 0/2 + 1/4 + 2/8 + 3/16 + 4/32 = sum((N-1)*(1/2^N)) for all N=1,2,… = 1.00.
    Hence, country’s b/g = 1.
    Why not higher? Because we have a 1/2 chance of terminating with 1 boy and 0 girls.

  2. Submitted By: Michael — November 13, 2007
    -2 votes
      + -

    An alternate explanation:

    The expected number of children a family will have is 1/P(b) where P(b) is the probability of having a boy. If P(b)=0.5, then each family is expected to have two kids. For a family to have two kids, the first must be a girl and the second a boy.

  3. Submitted By: J1g54w — June 5, 2008
    +0 votes
      + -

    e(g)=0*0.5+1*.25+. . .=1
    e(b)=0.5+0.25+. . .=1
    e(b)/e(g)=1

  4. Submitted By: venkateswara rao — June 22, 2008
    +1 votes
      + -

    ratio is 1/1

    explanation
    boys will be=1/2*1(first one is boy)+1/2*1/2*1(second one isboy)+(1/2)^3*1(third one is boy)……=(1/2)/(1-1/2)=1

    girls will be=1/2*0(first one is boy)+(1/2)^2*1(second one is boy)+(1/2)^2*2(third one is boy)…..=1

    proportion is 1/1

    note: p*n sum of products of probablity of having n (boys or girls) in a family is p

  5. Submitted By: sumanth — December 22, 2008
    not yet rated
      + -

    agree with the ans 1:1

  6. Submitted By: Christopher — February 25, 2009
    -1 votes
      + -

    Al, Exp[b/g] =NOT= Exp[b/g | 1 kid] + Exp[b/g | 2 kids] + Exp[b/g | 3 kids] + …

    Vekateswara, in your formulal should be
    0/2 + 1/4 + 2/8 + 3/16 + … (i-1)/(2^i)…
    although equal to 1/2 + 1/4 + 1/8 + 1/16 + … = 1
    it is not a “skip-line” calculation

    here is my proof

    !!! for X taking only non-negative integer values
    E[X] = Sum_{x>=1} of P[X >= x] !!!

    E[boys]
    = P[boys >= 1] + P[boys >= 2] + … = 1 + 0 + 0 + …
    = 1

    E[girls] =
    = Pr[girls >= 1] + Pr[girls >= 2] + … Pr[girls > i]… =
    = 1 /2 + 1/4 + … 1/(2^i)…
    = 1

    => E[b/g] = E[b] / E[g] = 1

  7. Submitted By: Anne — March 29, 2009
    +0 votes
      + -

    The calculations shown have not taken into account the fact that more boys are born than girls. The typical boy-girl birth ratio is 105 males born for every 100 girls. I’ll let you math whizzes figure out the rest!

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