For what value of $N$ those events are independent?

Question

There are $N$ children in a family.

The probability for a child to be a son is $1/2$.

We define $A$ - the family has boys and girls.

$B$ - the family has at most one girl.

For which values of $N$ the events $A$ and $B$ are independent?

Thank you

Answer 1

The converse event of $A$ is that the family has no girls or no boys. Thus

$P(A)=1-P(\overline A)=1-0.5^N-0.5^N=1-2\cdot 0.5^N$

The probability of event B is $P(B)= 0.5^N+N\cdot 0.5^N$

It is the probability of having no girls plus the probability of having one girl out of $N$ children.

Thus $P(A)\cdot P(B)=(0.5^N-2\cdot 0.5^{2N})\cdot (1+N)$

We have to show that $P(A)\cdot P(B)=P(A\cap B) $ for a specific value of $N$.

$P(A\cap B)=P(A)+P(B)-P(A \cup B)$

$P(A \cup B)$ is the probability that the family has not only boys. This is $1-0.5^N$

If $A$ and $B$ are independent then

$(0.5^N-2\cdot 0.5^{2N})\cdot (1+N)=1-2\cdot 0.5^N+0.5^N+N\cdot 0.5^N-1+0.5^N$

You can simplify this equation. After then try out some values for $N \in \mathbb N^+$