# Given the Probability, Find the Sample Space

Here is a simple but rather unusual problem:

A school teacher is in charge of a group of students. She wants to select two of them in random, and observes that it is exactly an even chance (50%) that they are of the same sex. What can be said about how many children of each sex there are?

### References

- Simon Norton,
__From Sex to Quadratic Forms__, An Invitation to Mathematics, D. Schleicher, M. Lackmann (eds), Springer, 2011

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Copyright © 1996-2017 Alexander Bogomolny

A school teacher is in charge of a group of students. She wants to select two of them in random, and observes that it is exactly an even chance (50%) that they are of the same sex. What can be said about how many children of each sex there are?

### Solution

From a group of *b* boys and *g* girls a teacher can form *b* + *g*)(*b* + *g* - 1)/2*bg* pairs are of different sexes. To insure the 50% chance of the same sex selection, the former number needs to be twice the latter:

(*b* + *g*)(*b* + *g* - 1)/2 = 2*bg*

which reduces to

(*b* - *g*)² = *b* + *g*.

Letting *b* - *g* = *n* leads to a system

*b* - *g* = *n*

*b* + *g* = *n*²

from which *b* = *n*(*n*+1)/2, *g* = *n*(*n*-1)/2. Cases where *n* = 0*n* = ±1*n* could be any integer. When *n* is negative, *g* > *b*;*b* > *g*.

In any event, *b* and *g* need to be two consecutive triangular numbers.

Rob Eastaway posted a modification on twitter:

Group of children, 3 of them boys. If I pick two children at random, there's a 50% chance both are boys. How many girls?

This is better be generalized. Let *T*_{n} = *n*(*n* + 1)/2 be the *n*-th triangular number. Then we can pose the follwing problem:

Group of children, *T*_{n} of them boys. If I pick two children at random, there's a 50% chance both are boys. How many girls?

Repteating the derivation above will leads to the answer *g* = *n*(*n* - 1)/2 = *T*_{n-1}.

To answer Rob's question, 3 = 2·3/2 = *T*_{2}, therefore, in his case, the number of girls is *T*_{1} = 1·2/2 = 1.

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