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A large community practices birth control in the following peculiar fashion. Each set of parents continues having children until a son is born; then they stop. What is the ratio of boys to girls in the community if, in the absence of birth control, 51% of the babies are born male?

  1. $51:49$
  2. $1:1$
  3. $49:51$
  4. $51:98$
  5. $98:51$
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Let, $X$ be the expected number of children a parent has.

So, expected number of boys $= 1$ and expected number of girls $= X - 1.$

The probability of having a baby boy $= 0.51.$

And the probability of having a baby girl $= 0.49.$

So, 

$X = 1 \times (0.51) + 2 \times (0.49) \times (0.51) + 3 \times (0.49)^2 \times (0.51) + 4 \times (0.49)^3 \times (0.51) $

$0.49X=                1 \times (0.49) \times (0.51) + 2 \times (0.49)^2 \times (0.51) + 3 \times (0.49)^3 \times (0.51) $


$X - 0.49X =   (0.51)[ 1 + (0.49) \times (0.51) + (0.49)^2 \times (0.51) +  (0.49)^3 \times (0.51) + \ldots ]$

$0.51X   = (0.51) [ 1 / 0.51 ]$

$\implies X  = 100 / 51.$

So, No of girl children $= X - 1.$

$\quad \quad = [100 / 51 ] - 1 = 49 / 51.$

No. of boy children $= 1.$

Hence, Ratio of Boys and Girls $= 51 : 49.$

[ Ans - A ]  

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