A school teacher is in charge of a group of students. She wants to select two of the 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?
A school teacher is in charge of a group of students. She wants to select two of the 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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