= Solution
Under the <null hypothesis> that treatment and recovery are independent, estimate each expected cell count by
$$
E_{ij}=\frac{(\text{row total})(\text{column total})}{N}
$$
and form the <Pearson chi-squared test of independence> statistic $\chi^2=\sum(O-E)^2/E$. For a $2\times2$ table its asymptotic null law is a <chi-squared distribution> with one <degree of freedom>. A second-order <Taylor expansion> of the <generalized likelihood-ratio test> statistic $-2\log\Lambda$ around the fitted counts gives this Pearson statistic, which explains its asymptotic null distribution.
Writing the four residuals over their fitted counts and simplifying gives
$$
\boxed{\chi^2=\frac{(ad-bc)^2(a+b+c+d)}{(a+b)(c+d)(a+c)(b+d)}.}
$$
For $(a,b,c,d)=(50,10,15,5)$,
$$
\chi^2=\frac{80}{117}\approx0.684,
$$
whose one-degree-of-freedom <p-value> is about $0.41$. There is no evidence of an effect. One fitted count is only $3.75$, however, so the chi-squared approximation is questionable; <Fisher's exact test> would be preferable.
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