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In a group of n people, two are randomly selected with probability 11/21 of being same gender. How many people are on the group if the number of males is twice as large as the number of females?

I've been trying to solve this problem by introducing a r.v X which represents the distribution of 2 males and of 2 females from M+F persons.

$$\frac{{M \choose 2}}{{M+F \choose 2}}+\frac{{F \choose 2}}{{M+F \choose 2}}=\frac{11}{21}$$

But how do I find n out of this?

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2 Answers

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$$ {\binom{2F}{2}+\binom{F}{2}\over\binom{3F}{2}}={2F(2F-1)+F(F-1)\over3F(3F-1)}={2(2F-1)+(F-1)\over3(3F-1)}={5F-3\over9F-3}={11\over21}\\ \therefore F=5,M=10 $$

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Let $x$ be the number of females. Then $2x$ is the number of males. The total population is $3x$. The probability of two males $$\Pr(MM) = \frac{2x}{3x} \frac{2x -1}{3x -1}=\frac{4x^2-2x}{9x^2-3x}.$$ Similarly, $$\Pr(FF)=\frac{x}{3x}\frac{x-1}{3x-1}=\frac{x^2-x}{9x^2-3x}.$$ The total probability should be $$\Pr(MM \cup FF) = \frac{4x^2-2x+x^2-x}{9x^2-3x}=\frac{5x^2-3x}{9x^2-3x}=\frac{11}{21}.$$ Solving for $x$ yields $x=5$, which means there are $5$ females, $2\cdot 5=10$ males, and $15$ people in total.

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