I have a problem understanding the result of this problem.
Can someone help me?
In a given week, in the gynecology department of a hospital $n$ women give birth. It may not occur multiple births, and boy or Girls' births are equally likely. In addition, it is assumed that the sex of the newborn is stochastically independent.
Consider $a_{n}$ the probability that at least 60% of newborns are girls.
I must prove that $lim_{n \rightarrow \infty} a_{n}=0$
The proposed solution is the following: Let $A_{j}$ be the event that at j-th birth a girl was born. Then $A_{1}... A_{n}$ are stochastically independent.
Let $X_{j}$ the number of girls among n births.
Then we have with $R_{n}:=\frac{1}{n}X_{n}$ following:
$a_{n}=P(X_{n}\geq n*0.6)\leq P(|R_{n}- \frac{1}{2}|\geq 0.1)$
With the weak law of large numbers the probability converges for $n \rightarrow \infty$ to zero.
What I cannot understand is why $P(X_{n}\geq n*0.6)$ is smaller than $P(|R_{n}- \frac{1}{2}|\geq 0.1)$ and where does the $0.1$ comes from.
Thank you for your help!
$X_n \geq (n)(0.6)$ implies $R_n \geq 0.6$. So $|R_n-0.5|\geq R_n-0.5\geq 0.6-0.5=0.1$. Finally, $A \subseteq B$ implies $P(A) \leq P(B)$.