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I'm working on a probability question:

Given the equiprobability of "having a boy" and "having a girl" as $1/2$ each, for what value of $n$ births, $n\geq2$ are the following two events independent?

event A: The family have two different sexes event B: Tha family have at most one girl 

I was able to narrow the equation down to $2^n - 2n - 2 =0$ for $n\geq2$ and knowing that $n=3$ is the solution, I still have to prove it.

1 Answers 1

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You can easily prove by induction that $2^n > 2n+2$ for all $n\geq 4$. This shows that there is no solution greater or equal to $4$.

P.S. This solution takes advantage of the fact that $n$ is integer. If you want to solve the equation in real numbers, let $f(x)=2^x-2x-2$.

Then

$f'(x)=2^x \ln(2)-2$

and it is easy to check that $f'(x) >0$ for all $x>2$. This implies that $f(x)$ is strictly increasing on $(2, \infty)$ and thus it has at most 1 solution.

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    Thanks! what was I thinking? I guess I wasn't thinking!2012-10-09