x amount of people owned a goat, y amount of people owned a camel, z amount of people had one animal or the other but not both In a survey, people were asked if they owned a goat or a camel. One person in fifteen said they had a goat. One person in eighteen said they had a camel and a tenth of the people had one animal or the other but not both. What proportion of the people owned neither kind of animal?
Proportion of people who had a goat = 1/15
Proportion of people who had a camel = 1/18
Proportion of people who had one animal or the other = 1/10
Proportion of people who owned neither kind of animal
= 1 - 1/15 - 1/18 - 1/10 = 7/9
I'm told however that the answer is actually, 8/9.
Any ideas on where my understanding is breaking down?
 A: Say $90$ people were surveyed (LCM of $18, 15$ and $10$).
So going by the ratios given, $6$ of them have goats ($G$), $5$  of them have camels ($C$) and $9$ of them have one of the animals but not both. So,
$|G| = 6, |C| = 5$
Please note that both the sets $G$ and $C$ also include people who have both animals.
So,
$(|G| - |G \cap C|) + (|C| - |G \cap C|) = 9$
$\implies |G \cap C| = 1$
So, $|G \cup C| = |G| + |C| - |G \cap C| = 10$
So out of $90$ people, $80$ of them have no animals, which is indeed $8/9$.
A: Write $G$ for owning a goat and $C$ for owning a camel.
Consider drawing a Venn Diagram and labelling the spaces inside the circles as $x, y$ and $z$, with the space outside the circles as $1-x-y-z$.
Let $p(G\cap C)=y$, $p(G\cap C')=x$ and $p(C\cap G')=z$.
Then from the data you have the following equations:
$$x+y=\frac{1}{15}$$
$$y+z=\frac{1}{18}$$and $$x+z=\frac{1}{10}$$
Solving this system of equations gives::$x=\frac{1}{18}$, $y=\frac{1}{90}$ and $z=\frac{2}{45}$
For the final answer you want $1-x-y-z$ which indeed works out to be $\frac89$
