Not just a Troll (TKC Submission)

Probability Level 4

$4$ children were born at Promila Hospital last week. Let us assume (according to the basic laws of Biology) that each child is equally likely to be a boy or a girl. What is the most probable ratio (gender) of the births?

Choose from the options below.

Note: This may seem like a troll question, but it isn't.
Anyway, I am submitting this as my entry in the The Troll King Competition. Kindly upvote it here if you liked it

2 of each gender All 4 are boys Meh, Any of these outcomes is possible All these outcomes are equally likely All 4 are girls Dummy option (This is the right answer) 3 are of 1 gender, 1 of the other

3 solutions

Ivan Koswara
Mar 22, 2015

First, we discard the unnecessary options, leaving us with:

All four are boys
All four are girls
Two of both genders
Three of one gender, one of the other
All are equally likely

We can compute the probability of each option.

All four boys: This is trivially $\left( \frac{1}{2} \right)^4 = \frac{1}{16}$ .
All four girls: Same thing, $\frac{1}{16}$ .
Two of both genders: This is $\binom{4}{2} \cdot \left( \frac{1}{2} \right)^2 \cdot \left( \frac{1}{2} \right)^2 = \frac{6}{16}$ .
Three of one gender, one of the other: This is the trap! This is the union of two options: three boys one girl, and three girls one boy. Each of these two possibilities have probability $\binom{4}{1} \cdot \left( \frac{1}{2} \right)^1 \cdot \left( \frac{1}{2} \right)^3 = \frac{4}{16}$ , but since there are two of them, we double it to give $\frac{8}{16}$ .

Thus three of one gender, one of the other is the most likely option.

A probability generating function (also referred to as binomial probability distribution) is the fastest way, in my opinion.

Denote by $m$ and $f$ the male and female gender respectively. Since $4$ babies were born, we have $4$ trials for the gender distribution and as the elementary events are equally likely, we have their probabilities as $\frac{1}{2}$ for each trial (birth). The probability generating function for this will be,

$\left(\frac{1}{2}m+\frac{1}{2}f\right)^4=\frac{1}{16}m^4+\frac{1}{4}m^3f+\frac{3}{8}m^2f^2+\frac{1}{4}mf^3+\frac{1}{16}f^4$

From this, it is trivial that the probability for $3$ of one gender and $1$ of the latter gender is the sum of the coefficients of $m^3f$ and $mf^3$ which is $\frac{1}{2}$ and is greater than the probabilities of all the other events of the options.

Prasun Biswas - 6 years, 2 months ago

Obtaining the probability generating function requires the exact same method used here (calculate each coefficient).

Ivan Koswara - 6 years, 2 months ago

https://brilliant.org/problems/too-many-puppies/

Agnishom Chattopadhyay - 6 years, 2 months ago

Oh my God! I seriously haven't seen this problem of yours! :O

Krishna Ar - 6 years, 2 months ago

No problem, dude!

Agnishom Chattopadhyay - 6 years, 2 months ago

What is there to get trolled in here ? A very straightforward problem in my opinion.

Venkata Karthik Bandaru - 6 years, 2 months ago

Some people will just answer "two and two are the most likely", interpreting "three and one" as three of one particular gender (say boys) and one of the other, thus only counting half of the possibilities.

Ivan Koswara - 6 years, 2 months ago

Haha, very funny way to get trolled !

Venkata Karthik Bandaru - 6 years, 2 months ago

@Venkata Karthik Bandaru – And it became a level 4 question.

Aryan Gaikwad - 6 years, 2 months ago

Tom Engelsman
Jan 2, 2016

Very easy problem using ordered-quadruples (i.e. (MMMM), MFMF), (FFFM), etc.). Why is this rated as Level-4?

Kuldeep Guha Mazumder
Aug 6, 2015

It is a very simple application of the binomial distribution.

Not just a Troll (TKC Submission)

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