How to get the size of the attempts in this exercise? How many children should a couple have so that, with probability $0.95$, she has at least one son and one daughter?


I've thinked the problem like a Binomial Distribution, Assuming that the probability of having a son is equal to having a daughter then $p=0.5$, if $X:= '\text{Number of sons obtained in n attempts}'$ then $X=B(n,0.5)$.

So, for find the $n$ i need:
$P(1\leq X \leq n-1)=0.95$

Because, i need that at least she have a daughter.

Am I right ?, or how should this exercise be solved?
Yes, you are right. Now calculate the sum and solve the inequality:
$$P(1\leq X \leq n-1)\ge 0.95 \iff \\
\sum_{k=1}^{n-1} {n\choose k}\left(\frac12\right)^k\left(\frac12\right)^{n-k}\ge 0.95 \iff \\
\frac1{2^n}\sum_{k=1}^{n-1} {n\choose k}\ge 0.95 \iff \\
\frac1{2^n}\left(2^n-2\right)\ge 0.95 \iff \\
0.05\cdot 2^n\ge 2 \iff \\
2^n\ge 40 \iff \\
n\ge 6. $$
Not sure this is right but with 50% chance boy or girl is 0.5
If you say the following ( Boy / Girl there is 50% chance each time )  then....
0.5 X 0.5 X 0.5 X 0.5 X 0.5 = 0.03125

the 3.125% here falls under the 5% margin between 1.00 and 0.95
so the answer could be 5 children.

However we all know nature can be unpredicatable and couples have been known to have lots of just boys or just girls.

Regards,

John O.