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?

Solution

References

  1. 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 pairs of which 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 = 2bg

which reduces to

(b - g)² = b + g.

Letting b - g = n leads to a system

b - g = n
b + g = n²

from which b = n(n+1)/2, g = n(n-1)/2. Cases where n = 0 (no kids at all), or n = ±1 (one sex is missing) can be disregarded. Other than those, n could be any integer. When n is negative, g > b; when it is positive, 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 Tn = n(n + 1)/2 be the n-th triangular number. Then we can pose the follwing problem:

Group of children, Tn 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 = Tn-1.

To answer Rob's question, 3 = 2·3/2 = T2, therefore, in his case, the number of girls is T1 = 1·2/2 = 1.

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Copyright © 1996-2017 Alexander Bogomolny

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