Question: > 
> Solve for $\bar{x}$ and $\bar{y}$  
> 
>  $$\bar{x}+\bar{y}=\bar{a},~~ \bar{x}\times \bar{y}=\bar{b},~~ \bar{x}.\bar{a}=1$$
> 
> 
> 

Attempt:

$\bar{x}+\bar{y}=\bar{a}$ dot by $\bar{a}$, we get $1+\bar{a}.\bar{y}=|\bar{a}|^2$ i.e $\bar{a}.\bar{y}=|\bar{a}|^2-1$

$\bar{x}\times \bar{y}=\bar{b}$ pre-cross by $\bar{a}$, we get $$(\bar{a}.\bar{y})\bar{x}-(\bar{a}.\bar{x})\bar{y}=\bar{a}\times \bar{b}\implies (|\bar{a}|^2-1)\bar{x}-\bar{y}=\bar{a}\times \bar{b}$$

I need another such equation to solve $\bar{x}$ and $\bar{y}$. Please help.

Answer: 
There is no real sense in formulating an expected error because you should say an expected error relative to some distribution, but you don't know what that distribution is. Nonetheless, the "standard" statistical answer here is as follows. The number of $1$s is distributed as Binomial($m,p$) where $p=1/n$ would be the unbiased case. This has mean $mp$ and variance $mp(1-p)$. From here you can construct a confidence interval for $p$ using the normal approximation to the binomial distribution. 

For instance if you observed $\hat{p}$ then the half-width of the confidence interval for $p$ at the level of significance $1-\alpha$ is $z\_{\alpha/2} \left ( \frac{\hat{p}(1-\hat{p})}{m} \right )^{1/2}$, where $z\_\beta$ satisfies $P(Z \geq z\_\beta)=\beta$ when $Z$ has the standard normal distribution. A common choice of $\alpha$ is $0.05$; in this case you have $z\_{0.025} \approx 1.96$. The significance of this is that approximately $95\%$ of intervals generated in this fashion will contain the true value $p$. Another way of interpreting the significance here is that you would accept the null hypothesis $p=p\_0$ at the 95% significance level if and only if $p\_0$ is in your confidence interval.

The *deficit* of this is that you have no way of actually knowing for sure whether the *particular* interval that *you* generate contains the true value $p$. All you can do is speculate about what would happen if you generated many such intervals (which we rarely do in the real world).
You can't measure the "expected error".

In probability, it is thought that if you made an experement N times(N should be a big number) and the interested eveniment happend "k" times, then the fraction ( k / N ) tends to an number supposed to be "the real probability".

So, by repeating the experimen many times, all you know is that "the bigger is N, the better your estimation is", and not "how close you are".

I don't know if you are able to do a few milions test, but that number should be enought. Sorry if this isn't the expected answer