I can't figure out how to set up a tree diagram of the possibilities and also how to set this problem (using Bayes' theorem) up to solve it. I've learned the basic formula for Bayes' Theorem, but not in more complex situations like this;
Three students, Lauren, Madi and Georgi, work in a coffee shop. In any given hour, Lauren will make 55% of the coffee, Madi will make 35% and Georgi will make 10%. The probability that Lauren, Madi, and Georgi spill some coffee is 0.6, 0.2, and 0.1, respectively.
Given that a coffee is spilled, find the probability that it was served by Madi.
On the most granular level the events are the event $\mathrm{LS}$ that Lauren serves the coffee and spills it, the event $\mathrm{LN}$ that Lauren serves and does not spill, the event $\mathrm{MS}$, $\mathrm{MN}$, $\mathrm{GS}$ and $\mathrm{GN}$, defined analogously. The probabilities are $P(\mathrm{LS})=0.55\cdot 0.6$, $P(\mathrm{LN})=0.55\cdot 0.4$, $P(\mathrm{MS})=0.35\cdot 0.2$, $P(\mathrm{MN})=0.35\cdot 0.8$, $P(\mathrm{GS})=0.1\cdot 0.1$, $P(\mathrm{GN})=0.55\cdot 0.9$. Now, you want the probability $P(\mathrm{M}|\mathrm{S})$ that the coffee was served by Madii given that it was spilled. Using the definition we have $P(\mathrm{M}|\mathrm{S})=\frac{P(\mathrm{MS})}{P(\mathrm{S})}$. The probability $P(\mathrm{S})$ that the coffee was spilled is $P(\mathrm{S})=P(\mathrm{LS})+P(\mathrm{MS})+P(\mathrm{GS})$ obtained by the respective sum of the granular events on which the coffee was spilled. Substituting the numbers, we get $P(M|S)=\frac{0.35\cdot 0.2}{0.55\cdot 0.6+0.35\cdot 0.2+0.1\cdot 0.1}$. Since I'm afraid to miscalculate, I omit the last step.