Probability - Taxi Question The probability of hiring a taxi from garage A, B and C is $0.3$, $0.5$ and $0.2$ respectively. The probability that the taxi ordered will be late from A is $0.07$, from B is $0.1$ and from  C is $0.2$.
(i) Illustrate this information on a tree diagram showing the probability on all branches. 
I have attempted this part of the questions and the tree diagram is attached in the link.
(ii) A garage is chosen at random, determine the probability that 
a) the taxi will arrive late 
I have attempted this as well: 
$$P (\text{Taxi arriving late}) = 0.07 + 0.1 + 0.2 = 0.37$$
Is this correct?
b) the taxi will come from Garage C given that it is late.
I'm not sure how to work this one. 

 A: For $(a)$ you have to weight the probabilities according to the probabilities of that taxi being selected.
We have
$$\mathsf P(\text{late})=0.3\cdot0.07+0.5\cdot0.1+0.2\cdot0.2=0.111$$
For $(b)$ use Bayes' Theorem:
$$\mathsf P(\text{C}\mid \text{late})=\frac{\mathsf P(\text{C}\cap\text{late})}{\mathsf P(\text{late})}$$
where $\mathsf P(\text{late})$ was obtained from part $(a)$
A: No, your answer to (a) is not correct. You want to find $$\begin{align}P(\text{taxi late})&=P(\text{taxi late}|A)P(A)+P(\text{taxi late}|B)P(B)+P(\text{taxi late}|C)P(C)\\&=(0.07\times0.3)+(0.1\times0.5)+(0.2\times0.2)\\&=\cdots\end{align}$$Here we used the law of total probability.
Hint for (b): $$P(C|\text{taxi late})=\frac{P(\text{taxi late}\cap C)}{P(\text{taxi late})}=\frac{0.2\times0.2}{\text{answer from part (a)}}$$
This is the definition of conditional probability.
A: The tree diagram is fine. The probability of being late is not correct.\begin{align}
P(\text{late}) &= \sum_{i \in \{A, B, C\}}\color{blue}{P(i)}P(\text{late}|i) 
\end{align}
You did not include the part that I highlighted.
For the last part, use the following:
\begin{align}
P(C|\text{late}) = \frac{P(C \cap \text{late})}{P(\text{late})} = \frac{P(C)P(C|\text{late})}{P(\text{late})}
\end{align}
