TIFR CSE 2014 | Part A | Question: 11

Question

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?

• A. $51:49$
• B. $1:1$
• C. $49:51$
• D. $51:98$
• E. $98:51$

Answer 1

(A) should be the correct choice.

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In the community we know that each set of parents will have exactly $1$ boy.

The number of girls might differ.

To find the ratio of boys to girls in the community, we are going to find the the expected number of girls that each parent set can have. 

Henceforth in this question we are going to use the word "family" to denote a "parent set". 

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Calculation of expected number of girls in any family 

Let $X$ be a random variable that denotes the number of girls that any family. Each family will have exactly $1$ boy. 

$P\left( X = 0\right)$ will denote : Probability that a family has $0$ girls and $1$ boy.

$P\left( X = 0\right) = \left( 0.49\right)^{0}\left( 0.51\right) $ 

$P\left( X = 3\right)$  will denote : Probability that a family has $3$ girls and $1$ boy.

$P\left( X = 3\right) = \left( 0.49\right)^{3}\left( 0.51\right)$

and so on.

In general we can say that

$P\left( X = i\right)$  will denote the probability that a family has $i$ girls and $1$ boy, and

$P\left( X = i\right) = \left( 0.49\right)^{i}\left( 0.51\right)$ .

Now the expected number of girls in any family will be denoted by $E\left[X\right]$.

Here,

$E\left[X\right] = \sum_{i=0}^{\infty} i\cdot P\left( X = i \right)$.

but $P\left( X = i\right) = \left( 0.49\right)^{i}\left( 0.51\right)$. So, we get,

$E\left[X\right] = \sum_{i=0}^{\infty} i\cdot \left( 0.49\right)^{i}\left( 0.51\right)$.

This implies $E\left[X\right] = \left( 0.51\right)\sum_{i=0}^{\infty} i\cdot \left( 0.49\right)^{i}$.

The formula for summation of series of type $\sum_{k=0}^{\infty} k\cdot x^{k}$ can be found by differentiating

$\sum_{k = 0}^{\infty}x^k$ with respect to $x$.(See the reference below).

This gives

$\sum_{k=0}^{\infty} k\cdot x^{k} = \frac{x}{\left(x-1\right)^2}$

So, $\sum_{i=0}^{\infty} i\cdot \left(0.49\right)^{i} = \frac{0.49}{\left(0.49-1\right)^2}$

Hence,

$E\left[X\right] = \left( 0.51\right)\cdot \frac{0.49}{\left(0.51\right)^2} = \frac{0.49}{0.51}$

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Now the ratio of boys to girls can be given by number of boys in each family/expected number of girls in each family.

i.e., $Ratio\left(\text{B to G}\right) = \frac{1}{E\left[X\right]}$

So, $Ratio\left(\text{B to G}\right) = \frac{1}{\frac{0.49}{0.51}} = \frac{51}{49}$

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Reference for series summation: 

http://math.stackexchange.com/questions/629589/converge-of-the-sum-sum-k-1n-k-xk

Comments

"In the absence of birth control, 51% babies of the babies are born male."

It means that any new born baby in the community is more likely to be a boy, rather than a girl.

It does not means that community is not following that birth control.It is following birth control but in case it was not following then the ratio would have been 51/49.

Coincidentally, here ratios are same in both the cases.

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one other way to calculate this series

S=1.k¹+2.k²+3.k³+4.k⁴+5.k⁵+.........

kS=        1.k²+2.k³+3.k⁴+4.k⁵+..............

S-kS= k+k²+k³+k⁴+k⁵+k⁶..........

S(1-k)=k/1-k

so S=k/(1-k)²

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Why don’t we consider the parents themselves in the final calculation?
The whole population would involve them too, if we don’t implicitly assume that they both pass away as soon as a boy is born in the family.
For that calculation though, the numerator and denominator would have 2 +1’s, each for father and mother, and final aswer would be $ \frac{51}{50}$. This, not being present in the options; there is indeed the implicit assumption that the parents aren’t part of the population.

Also I wonder, how the calculation for a single parent could be taken as the average for all, especially since their own children could be parents themselves. So, a lot of implicit assumptions here. Be VERY careful.

Answer 2

Alternative approach to solve:

Let us look at  the event when the number of children(C) born(until we get a son)  is 1, 2, 3, 4, 5,……,N.

S → Representation of the birth of son.

D → Representation of the birth of daughter.

When C=1, then only son is born i.e. S .

C=2 , then one daughter and one son is born i.e. D S .

C=3, then two daughter and one son is born i.e D D S.

……………………………. this goes on till N

So when C =N, then (N-1) daughters and one son is born i.e. D D D D …...(N-1 times)   S .

The total number of sons = N .

The total number of daughters = 1 + 2 + 3 + 4 + ...(N-1)  =$N(N-1 )/2$ .

So total number of children = Total number of sons + Total number of daughters = N + N(N-1)/2 = N(N+1)/2.

It is given that 51% of the children are males.

So the equation that we get is

Total number of sons / Total number of children = 51/100

i.e.  $\frac{N}{N(N+1)/2 } = \frac{51}{100}$

so N = $\frac{149}{51}$

The Ratio of Number of son to Number of daughters = $N : \frac{N(N-1)}{2}$

                                                                                   = 51 : 49

Answer 3

Let  the total population be T. Assuming the population consist of only boys and girls,

probablity of  boys= 51/100

probablity of girl= 49/100

number of boys in the population= B= 51T/100

number of girls in the population = G= 49T/100

B/G= 51/49   hence option A is correct