I am going to assume that "100% survival" means that your sites only contained a single organism. so 30 means 30 organisms died, and 31 means 31 organisms didn't.
Based on this the chi-square should be fine, but it will only tell which hypothesis are not supported by the data - it won't tell you if two reasonable hypothesis are better or not. I present a probability analysis which does extract this information - it agrees with the chi-square test, but it gives you more information than the chi-square test, and a better way to present the results.
The model is a bernouli model for the indicator of "death", $Y_{ij}\sim Bin(1,\theta_{ij})$ ($i$ denotes the cell of the $2\times 3$ table, and $j$ denotes the individual unit within the cell).
There are two global assumption underlying the chi-square test:
- within a given cell of the table, the $\theta_{ij}$ are all equal, that is $\theta_{ij}=\theta_{ik}=\theta_{i}$
- the $Y_{ij}$ are statistically independent, given $\theta_{i}$. This means that the probability parameters tell you everything about $Y_{ij}$ - all other information is irrelevant if you know $\theta_{i}$
Denote $X_{i}$ as the sum of $Y_{ij}$, (so $X_{1}=30,X_{2}=10,X_{3}=1$) and let $N_{i}$ be the group size (so $N_{1}=61,N_{2}=30,N_{3}=11$). Now we have a hypothesis to test:
$$H_{A}:\theta_{1}=\theta_{2},\theta_{1}=\theta_{3},\theta_{2}=\theta_{3}$$
But what are the alternatives? I would say the other possible combinations of equal or not equal.
$$H_{B1}:\theta_{1}\neq\theta_{2},\theta_{1}\neq\theta_{3},\theta_{2}=\theta_{3}$$
$$H_{B2}:\theta_{1}\neq\theta_{2},\theta_{1}=\theta_{3},\theta_{2}\neq\theta_{3}$$
$$H_{B3}:\theta_{1}=\theta_{2},\theta_{1}\neq\theta_{3},\theta_{2}\neq\theta_{3}$$
$$H_{C}:\theta_{1}\neq\theta_{2},\theta_{1}\neq\theta_{3},\theta_{2}\neq\theta_{3}$$
One of these hypothesis has to be true, given the "global" assumptions above. But note that none of these specify specific values for the rates - so they must be integrated out. Now given that $H_{A}$ is true, we only have one parameter (because all are equal), and the uniform prior is a conservative choice, denote this and the global assumptions by $I_{0}$. so we have:
$$P(X_{1},X_{2},X_{3}|N_{1},N_{2},N_{3},H_{A},I_{0})=\int_{0}^{1}P(X_{1},X_{2},X_{3},\theta|N_{1},N_{2},N_{3},H_{A},I_{0})d\theta$$
$$={N_{1} \choose X_{1}}{N_{2} \choose X_{2}}{N_{3} \choose X_{3}}\int_{0}^{1}\theta^{X_{1}+X_{2}+X_{3}}(1-\theta)^{N_{1}+N_{2}+N_{3}-X_{1}-X_{2}-X_{3}}d\theta$$
$$=\frac{{N_{1} \choose X_{1}}{N_{2} \choose X_{2}}{N_{3} \choose X_{3}}}{(N_{1}+N_{2}+N_{3}+1){N_{1}+N_{2}+N_{3} \choose X_{1}+X_{2}+X_{3}}}$$
Which is a hypergeometric distribution divided by a constant. Similarly for $H_{B1}$ we will have:
$$P(X_{1},X_{2},X_{3}|N_{1},N_{2},N_{3},H_{B1},I_{0})=\int_{0}^{1}P(X_{1},X_{2},X_{3},\theta_{1}\theta_{2}|N_{1},N_{2},N_{3},H_{B1},I_{0})d\theta_{1}d\theta_{2}$$
$$=\frac{{N_{2} \choose X_{2}}{N_{3} \choose X_{3}}}{(N_{1}+1)(N_{2}+N_{3}+1){N_{2}+N_{3} \choose X_{2}+X_{3}}}$$
You can see the pattern for the others. We can calculate the odds for say $H_{A}\;vs\;H_{B1}$ by simply dividing the above two expressions. The answer is about $4$, which means the data support $H_{A}$ over $H_{B1}$ by about a factor of $4$ - fairly weak evidence in favour of equal rates. The other probabilities are given below.
$$\begin{array}{c|c}
Hypothesis & probability \\ \hline
(H_{A}|D) & 0.018982265 \\
(H_{B1}|D) & 0.004790669 \\
(H_{B2}|D) & 0.051620022 \\
(H_{B3}|D) & 0.484155874 \\
(H_{C}|D) & 0.440451171 \\
\end{array}
$$
This is showing strong evidence against equal rates, but not in strong evidence favour of a defintie alternative. It seems like there is strong evidence that the "offshore" rate is different to the other two rates, but inconclusive evidence as to whether "inshore" and "mid-channel" rates differ. This is what the chi-square test won't tell you - it only tells you that hypothesis $A$ is "crap" but not what alternative to put in its place
