P(bicycle having lights infront and back) = 0.31
P(bicycle only having lights infront) = 0.23
P(bicycle only having lights in the back) = 0.09.
The rest of the bicycles had no lights.
If you know that a bike has working frontlights, what is the probability that it also has working backlights?
I tried using conditional probability, the chance of backlights working given that it has working frontlights: $P(B|A) = \frac{P(A \cup B)}{P(A)}$ but I cant find a way to use this correctly to get the right answer. Any ideas?
$$\begin{align}P(\text{Bike has working backlights }|\text{ Bike has working frontlights}) \\= \frac{P(\text{Bike has working backlights and frontlights})}{P(\text{Bike has working frontlights})} =\frac{0.31}{0.23+0.31} &=0.57\end{align}$$
This follows from $$P(B|A) = \frac{P(B\cap A)}{P(A)}$$