Bayesian probability problem? Problem: In a city there are three types of taxis which drive towards the airport. 30% are blue, 20% green, 50% yellow. They take there customers too late with probabilities 0.1,0.2,0.3 respectively. One day a passenger takes a taxi and tells the driver to drive to the airport. Once arriving, the passenger was not late and took the plane on time. What is the probability that he passenger drove in the yellow cab/taxi.
I attempted to solve the problem using the following approach:

But the result does not match one of the given results.

 A: 
$$\frac{0.5\times 0.7}{0.3\times 0.9+0.2\times 0.8+0.5\times 0.7}=\frac{0.35}{0.78}=\frac{35}{78} $$
A: Outline: It is a standard conditional probability / Bayes' Theorem calculation. 
Use the various probabilities of being on time, that is, $0.9$, $0.8$, and $0.7$.
Remark: In general, $\Pr(A\mid B)\ne 1-\Pr(A\mid B')$.  For an extreme example, let $B$ be some event that has non-zero probability, and let $A$ be an event that has probability $0$.
A: You want to find the probability that the taxi was yellow given that the passenger was not late. Let $P(B)$=the probability that the passenger was not late (on time) and $P(A)$=the probability that the taxi was yellow. So the probability of A given B can be found by Bayes' theorem, which is generalized as $P(A|B)=\frac{P(A \cap B )}{P(B)}$. $P(A \cap B )$ is the intersection of of $P(A)$ and $P(B)$, or the probabilities that the cab is yellow and that the passenger was not late, which is $.5*(1-.3)=.35$, because the probability of a yellow cab is .5 and the probability of a yellow cab being late is .3. $P(B)$ is the probability that a cab was on time. Therefore, you get $P(B)=1-(.3*.5+.2*.2+.1*.3)=.78$. So you have $35/78$ as your final answer. 
A: It can help to identify what you are given and what you want, symbolically, then apply Bayes' Rule.

Problem: In a city there are three types of taxis which drive towards the airport. 30% are blue, 20% green, 50% yellow. They take there customers too late with probabilities 0.1,0.2,0.3 respectively. 

You are given that: $$\begin{array}{cc}\mathsf P(B)=0.30 & \mathsf P(L\mid B)=0.10\\ \mathsf P(G)=0.20 & \mathsf P(L\mid G)=0.20\\ \mathsf P(Y)=0.50 & \mathsf P(L\mid Y)=0.30\end{array}$$

Once arriving, the passenger was not late and took the plane on time. What is the probability that he passenger drove in the yellow cab/taxi.

You want to know: $\mathsf P(Y\mid L^\complement)$  ... which is not $1-\mathsf P(Y\mid L)$ as you seemed to be attempting.   Rather we use $\mathsf P(L^\complement\mid Y) = 1-\mathsf P(L\mid Y)$ and so forth:
$$\begin{align}\mathsf P(Y\mid L^\complement) =&~ \dfrac{\mathsf P(Y)\,\mathsf P(L^\complement\mid Y)}{\mathsf P(B)\,\mathsf P(L^\complement\mid B)+\mathsf P(G)\,\mathsf P(L^\complement\mid G)+\mathsf P(Y)\,\mathsf P(L^\complement\mid Y)}\\[1ex] =&~ \dfrac{0.50\,(1-0.30)}{0.30\,(1-0.10)+0.20\,(1-0.20)+0.50\,(1-0.30)}\\[1ex]=&~ \dfrac{35}{78}\end{align}$$
