On the way home from work Chris goes through a traffic light then passes over a level crossing, the probability that Chris stops at a traffic light is $\frac{2}{3}$ while the probability that he is stopped at the level crossing further up the road $\frac{1}{5}$.
(i) Represent this information in a tree diagram
(ii) Find the probability that Chris gets stopped once on his way home
(iii) Find the probability that Chris gets home without stopping
Here is my tree diagram:
S = Stops at Traffic light
G = Goes through traffic light
S1 = Stops at level Crossing
G1 = Goes through level Crossing
(ii) I'm stuck on this part, so he either, goes through the traffic light then stops at the level crossing or stops at the traffic light then goes through the level crossing.
So I'm assuming it's this:
$$P(S∩G1) + P(G∩S1))$$
Since it's one or the other, hence:
$$(\frac{2}{3}.\frac{4}{5})+(\frac{1}{3}.\frac{1}{5})$$
$$\frac{3}{5}$$
(iii) I used Baye's rule for this:
$$\frac{P(G∩G1)}{P(S∩G1)+P(G∩G1)}$$
Which is:
$$\frac{\frac{1}{3}.\frac{4}{5}}{(\frac{2}{3}.\frac{4}{5})+(\frac{1}{3}.\frac{4}{5})}$$
$$\frac{1}{3}$$
I have a strong feeling this is all incorrect, especially the 3rd part, if someone can help I would be grateful
ii) Is actually correct.
iii) Is simply $\frac{1}{3}.\frac{4}{5}$ so $P(G\cap G1)$.