Probablity of Boy

Question

I faced this question in a coding contest.

The ratio of boys to girls for babies born in Russia is 1.09 : 1 . If there is 1 child born per birth, what proportion of Russian families with exactly 6 children will have at least 3 boys?

I am new to probability. Though I have studied about binomial distribution just now but still not able to lead with this question. Can anybody help with detailed answer.

Answer 1

Assuming the question is what proportion of Russian families with exactly $n$ children will have at least $k$ boys?

The number of boys in a russian family of $n$ kids has a Binomial distribution of parameters $n$ and $p=\frac{1.09}{2.09}$

Which means that they will have exactly $k$ children with probability

$C^k_np^k(1-p)^{n-k}$

Thus the probability of having at least $k$ children is :

$\sum_{i=k}^nC^i_np^i(1-p)^{n-i}=1-\sum_{i=0}^{k-1}C^i_np^i(1-p)^{n-i}$

There is, to my knowledge, no exact simplification of this formula

Well, just replace $n$ with 6 and $k$ with 3.

Answer 2

We are trying to arrange the permutations of BBBGGG (b for boy, g for girl). This is so that we can find all ways a family of 6 can have 3 boys ad 3 girls in some order.

First, there is a $(\frac{1.09}{2.09})^3(\frac{1}{2.09})^3$ chance that this situation even arises. But there are $\binom{6}{3}$ ways to arrange the given sequence, so the answer is just $\boxed{\binom{6}{3}(\frac{1.09}{2.09})^3(\frac{1}{2.09})^3}$ which is around 31.1%.