Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 7 down vote accepted

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.

share|improve this answer
    
Thanks! what was I thinking? I guess I wasn't thinking! –  user31280 Oct 9 '12 at 2:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.