There are $2$ maternity hospitals in the town. In the first hospital $50$ children are born every day, in the second $100$ children.
Which hospital will most often experience that the number of new-born boys and new-born girls are the same? Why?
There are $2$ maternity hospitals in the town. In the first hospital $50$ children are born every day, in the second $100$ children.
Which hospital will most often experience that the number of new-born boys and new-born girls are the same? Why?
Make the usual (false) assumptions of independence, and equality of probability of girl and boy.
The probability that the smaller hospital hospital has an equal number of girls and boys is $\binom{50}{25}\left(\frac{1}{2}\right)^{50}.$
In the bigger hospital, the probability is
$\binom{100}{50}\left(\frac{1}{2}\right)^{100}.$
We want to compare these rather complicated numbers. The Stirling approximation is overkill, but it yields that the ratio of the first number to the second is about $\sqrt{2}$. (This comes from $\sqrt{100/50}$.) So not only do we know that the probability of equality is greater in the smaller hospital, we also know by roughly how much.
We can do it with less machinery. The idea is to show that if we increase the number of births by $2$, the probability of equality decreases.
With $2n$ births, the probability of equality is $\binom{2n}{n}\left(\frac{1}{2}\right)^{2n}.$ With $2n+2$ births, the probability is $\binom{2n+2}{n+1}\left(\frac{1}{2}\right)^{2n+2}.$ Calculate the ratio. By expressing the binomial coefficients in terms of factorials, we get a lot of cancellation. The ratio turns out to be $\frac{4(n+1)(n+1)}{(2n+2)(2n+1)},$ which simplifies to $1+\dfrac{1}{2n+1}$, a number $\gt 1$.
Remark: If you want to proceed much more informally, calculate the probability of equality of sexes in $2$ births, $4$, $6$, $8$, and $10$. There will be a steady decrease.
A surprising number of people believe that the probability of equality of heads and tails increases as the number of tosses gets large. Completely false!