So heres my working out for the question.
I figured that I need to find three probabilities: P(c|x), P(y|c) and P(z'|c) as it is the probability they ate contaminated food given they have symptoms X and Y but not Z. P(c|x) = P(x|c)*P(c)/P(x). However I do not have P(x) so I went about finding it (also needed P(y) and P(z)).
So I did P(x|c) = P(x AND c)/P(c) to find P(x AND c) and then I did P(y|c') = P(x AND 'c)/P(c) to find P(x AND c'). I figured P(x AND c) + P(x AND c') = P(X). I did a similar thing for P(Y) and P(Z) getting values:
P(X) = 0.0575 P(Y) = 0.104 P(Z) = 0.00399
After plugging back into the original questions, the answer seemed way too small (0.02). Where am I going wrong?
We want the conditional probability they ate a contaminated item. This is the probability they ate a contaminated item and consequently developed the symptoms over the probability that they developed the symptoms. The numerator should be straightforward, and the denominator requires the Law of Total Probability. Thus we get $$= \frac{P(C)\cdot P(X \cap Y \cap Z' | C)}{P(C)\cdot P(X \cap Y \cap Z' | C) + P(C') \cdot P(X \cap Y \cap Z' | C') }$$ By independence of the symptoms, we get $P(X \cap Y \cap Z' | C) = P(X|C)P(Y|C)P(Z'|C)$, and similarly for $C'$. Then we can plug in: $$= \frac{0.01 \cdot (0.8 \cdot 0.5 \cdot 0.7)}{0.01 \cdot (0.8 \cdot 0.5 \cdot 0.7) + 0.99 \cdot (0.05 \cdot 0.1 \cdot 0.999) }$$ and calculate to get 0.362