How to assess the strength of the acid?
Let's start with $\ce{HXO_3}$.

In $\ce{HXO_3}$, after removal of proton, an $\ce{XO_3-}$ ion is formed which is resonance stabilised. For example, if X was $\ce{Cl}$,



Now, this shows that the $X$ needs to have a tendency to pull the electrons towards itself which decreases in the order :$\ce{Cl&gt;Br&gt;I}$. So more resonance is present in $\ce{ClO_3^-}$ than in $\ce{BrO_3^-}$ making the conjugate base of $\ce{HClO_3}$ more stable. Hence it is more acidic. 
So acidic strength decreases as: $\ce{HClO_3&gt;HBrO_3&gt;HIO_3}$

Now let's compare $\ce{HXO_3}$ and $\ce{HXO_2}$. You should now know that the acidic nature depends on the stability of the resulting conjugate base. Here, the number of resonance structure for $\ce{HXO_3}$ is more. Hence, the acid strength decreases in the order:   $\ce{HClO_3&gt;HClO_2&gt;HClO}$. So its actually the number of resonance structures that affects the acidic strength.

Finally, let's come to hydrides of 16th group. Let me represent any 16th group element as $Y$. Now, as size of $Y$ increases, the $\ce{H-Y}$ bond length increases. This means, the $\ce{H-Y}$ bond becomes weaker and removal of $\ce{H+}$ ion becomes easier. The size of atom decreases in the order :$\ce{O&gt;S&gt;SE&gt;Te}$.

Hence, the acidic strength decreases as : $\ce{H_2Te&gt;H_2Se&gt;H_2S&gt;H_2O}$ 

This shows that electronegative is not the only reason. For example, if you considered $\ce{HF}$ and $\ce{HCl}$, then you may think that $\ce{HF}$ is more acidic because $\ce{F}$ is more electronegative and hence will be happy to get a negative charge. But, charge on a small atom is unfavourable. As atomic size increases, the charge can spread over a larger area. Hence, $\ce{Cl-}$ is more stable. 

The same trend is seen in case of hydrides of 16th group. The charge in $\ce{Se-}$ is spread over a larger area. Hence, its more stable.