Birth Certificate!

If there were only 2 deliveries, the probability of equal number of boys and girls would be 0.5.

If there were 200 deliveries, then for getting a half probability, we will need 100 and 99 boys and girls before the last one, and its probability will definitely be less than half.

So we get the general case- More the number of deliveries, lesser the chance of equal number of bays and girls. And so is the answer- Hospital A.

If boys and girls are born with identical (independent) probability $\frac{1}{2}$ , then there are exactly $\binom{n}{n/2}$ possible birth orders of boys and girls in a hospital in which $n$ total children are born. (To see this, note that it suffices to determine which children are boys -- all the others must be girls.)

Meanwhile, there are $2^n$ possible birth orders in total, all of which arise with equal probability.

Thus, the probability that a hospital with $n$ births has an equal number of boys and girls is $p(n):=\frac{\binom{n}{n/2}}{2^n}.$

We can explicitly compute that $p(200)=\frac{11318564332012910145675522134685520484313073709426667105165}{200867 255532373784442745261542645325315275374222849104412672}\approx 0.0563485,$ and $p(400)=\frac{1286906251692680412162198504005024834465136567263470602935613244699 0416546581494824792074439872061618945506074688346765}{3227812347608635 7370698989650037648429121322410365293910383241956758095275210514932870 5669160017228929487896496593436672}\approx 0.0398693<p(200).$

Alternatively, we can observe that the derivative of $p(n)$ is $p'(n)=-2^{-n} \binom{n}{n/2} \left(H_{n/2}-H_n+\log (2)\right),$ where $H_k$ is the $k$ -th harmonic number. As $H_{n/2}-H_n$ (which is negative) is bounded below by $-\log(2)$ , we see that $p'(n)<0$ -- so that $p(n)$ is decreasing -- for $n\geq 0$ ; whence we must have $p(200)>p(400)$ .

You may tempt to give the equal probability of having equal number of boys and girls for both hospitals viz. A and B, as the probability of having a boy and a girl is equal but it differs as the number of possible outcomes increases.

Let me generalize it this way. For example we have 2 babies born. The probability of delivering 1 boy and 1 girl is: a) First boy and second girl OR b) First girl and second boy

So it is: (1/2 * 1/2) + (1/2 * 1/2) = 1/2

Let the number of babies born is 4. Therefore, the probability of delivering 2 boys and 2 girls is a) First boy and second boy and third girl and fourth girl OR

b) First boy and second girl and third boy and fourth girl OR

c) First boy and second girl and third girl and fourth boy OR

d) First girl and second girl and third boy and fourth boy OR

e) First girl and second boy and third girl and fourth boy OR

f) First girl and second boy and third boy and fourth girl OR

So it is : (1/2 * 1/2 * 1/2 * 1/2) + (1/2 * 1/2 * 1/2 * 1/2) + (1/2 * 1/2 * 1/2 * 1/2) + (1/2 * 1/2 * 1/2 * 1/2) + (1/2 * 1/2 * 1/2 * 1/2) + (1/2 * 1/2 * 1/2 * 1/2) = 6/16 = 3/8

i.e, 4 !/(2! * 2!) is the favorable outcome and 2^4 is possible outcome

= 6/16.

So for 200 it will come as (200!/(100! * 100!) / 2^200 and for 400 it will come as (400!/(200! * 200!) / 2^400)

Therefore if the number of babies born are less the probability of having equal boys and equal girls is greater.

Thanks

P(A) = (.5^100)(.5^100) = 0.5^200 P(B) = (.5^200)(.5^200) = 0.5^400 Hence, P(A) > P(B)

you can use pascals triangle to solve this and examine the middle term divided by the sum. For 2, its 1 2 1. 2/4=.5; For 4, its 1 4 6 4 1. 6/16=.375; For 6, its 1 6 15 20 15 6 1. 20/64= .3125. So by this logic, f(200) is less than f(100).

Why everyone are so complicated T_T. Hospital A has 1/100 chance of having equally boys and girls while Hospital B has 1/200 chance :. It's very easy :\

Probability tends to more true when the sample space large so , hospital B has greater sample space than hospital A, So hospital B has greater chance.

Funny how I, without complex A-level maths, managed to solve this using intuition only. I felt that the odds are, realistically, never even. So, the less the number of total births, the less likely it is that there are outstanding uneven numbers of boys and girls. Therefore it is A. Given what was just said though, I did ignore the bit that said the chances were exactly even, because that gave two options as an answer, which of course means by my logic one should probably skip over that detail somewhat

As hospital B has more babies, the chances to have equal number of boys and girls are hardest.

Probability p = favourable cases/Total no. of cases. Here, favourable case is just one, 50% boys and 50% girls. Total cases include: 400 boys: 0 girls, 399 boys: 1 girl.. and so on. So total number of cases are 400 and 200. Therefore, pA = 1/200, pB=1/400

its nothing maths in it just use ur iq. out of 200 babies probability will be more of equal no. of babies than in 400 if u have 2 balls and someone else has 5 balls have more probability of having the balls of same color

If we have 400 babies, there are 401 possible combinations of babies:

--- 0 boy, 400 girls

--- 1 boy, 399 girls

--- 2 boys, 398 girls

..........

--- 400 boys, 0 girls

And in only one case, the numbers are equal(200,200). So, the chance of it happening, for 400 babies, is 1/401.

If the same thinking is applied to the other hospital, we come to a probability of 1/201.

As 1/201>1/401, probability of hospital A is bigger than hospital B.

In hospital B of 400 babies, probability will be half that of hospital A of 200 babies. Thus Hospital A has more chance!

The number of ways to produce n of 2n boys is $\binom{2n}{n}$ . By Stirling's formula, this is approximately $\frac{4^n}{\sqrt{\pi n}}$ . Thus the probability of having exactly half boys half girls is $\frac{4^n}{\sqrt{\pi n}} / 2^{2n} = \frac{1}{\sqrt{ \pi n}}$ . Thus the greater the number of trials, the more difficult it is to get exactly half.

It is simple as the population size increases the chances if reaching the universal average increases. That is why hospital A with small number of babies has a higher chance of diverting from the universal average and having equal number if boys and girls.