Proof of symmetric probability involving a balanced coin

Question

A balanced coin is tossed $280$ times. Let $X$ be the number of H’s in the first 200 tosses, and let $Y$ be the number of H’s in the last 80 tosses. Prove that $P(X-Y≤50)≤7/20$

My attempt: Let’s mark $Z=X-Y$. The way I see it, we need to use here Cebyshev’s inequality. I found that $E(Z)=60 , V(Z)=70$. Now, I want to prove that $P(Z≥70)=P(Z≤50)$, because then I’ll get $P(Z-E(Z)≥10)=P(E(Z)-Z ≥10)$ , and then :

\begin{align*} P(Z\leq 50) &= P(Z-E(Z)\leq -10) \\ &= 0.5 ( P(Z-E(Z)\leq-10) + P(E(Z)-Z \geq10) ) \\ &= 0.5 (P(Z-E(Z)\leq-10) + P(Z-E(Z)\geq10) ) \\ &=0.5 P( |Z-E(Z)|\geq 10) \end{align*}

And from here we can use Cebyshev’s inequality and get the right result. The problem is that I don’t succeed of proving that $P(-Z≥-50)=P(Z≥50)$. In order to do that, I want to prove that $Z\sim N(60,70)$ (from here it will be easy to show what I want)- but I don’t see how. Do you know how can I prove it? Or maybe there is another way?

Answer 1

I was always bad with Chebyshev, so let me try a direct proof. Your question is related to this one, especially answer by Robert Israel. Wiki about binomial also helps (sum of two binomials with the same probability is binomial).

Your $X\sim\mathcal{B}(200,\frac{1}{2})$ and $Y\sim\mathcal{B}(80,\frac{1}{2})$. Hence $80-Y\sim\mathcal{B}(80,\frac{1}{2})$ and thus $X+80-Y\sim\mathcal{B}(280,\frac{1}{2})$. Thus $\mathbb{P}[X-Y\geq70]=\mathbb{P}[X-Y+80\geq150]$ and $\mathbb{P}[X-Y\leq50]=\mathbb{P}[X-Y+80\leq130]$ since $\mathcal{B}(280,\frac{1}{2})$ is symmetric around $140$.

Answer 2

$Z$ is not normally distributed so you can't prove it that way.

However, you don't need to show that it is $N(60,70).$ You just need to show that the distribution is symmetric about $60,$ or, in other words that the variable $X-Y-60$ is symmetric about zero.

We can write $X-Y-60 = X'-Y'$ where $X'=X-100$ and $Y'=Y-40.$ Now $X'$ and $Y'$ are mean zero, and a little thought about the binomial shows they are symmetric about zero. So $-X'$ has the same distribution as $X$ and $-Y'$ has the same distribution as $Y$ (and $X'$ and $Y'$ are independent) and thus $X'-Y'$ has the same distribution as $Y'-X'$ so $X'-Y'$ is symmetric about zero. Thus $X-Y$ is symmetric about $60$.