Statistics Uniform Distributuon Help

Antonio arrives at a train station at a random point in time. The trains to his desired destination are scheduled to depart at 12-minute intervals.
a) Assume that Antonio gets on the next train.
i) Suggest an appropriate distribution to model his waiting time and give the
parameters.
ii) State the mean and the variance of this distribution.
iii) State an assumption you have made in suggesting this distribution. [4]
b) Now assume that the probability that Antonio misses the next available train
because he is distracted by his smartphone is 0.12. If he misses the next available
train, he is sure to get on the one after that.
i) Find the probability that he waits between 9 and 19 minutes.
ii) Given that he waits between 9 and 19 minutes, find the probability that he
gets on the first train.

I understand the first part of the question, what I don't really understand is the part b. Can anyone help?

They'll wait between 9 and 19 minutes if they arrive <= 3 min and catch the first train OR arrive >= 5 min and miss the first train.

the solution for the question adds the probability of P(9<x<12) and the probability of P(12<x<19) together, I want to understand why they did that

P(9<x<12) will represent arriving <= 3 and catching the first triam and P(12<x<19) represents arriving >= 5 min and missing the first train. x will represent wait time.

Note it helps to upload your working/the ms youre looking at when you ask the question.

Thank you, I've uploaded the marking scheme. So the first one is implying he does not miss the first train and just waits three minutes, then the second is implying he does miss the first train and waits seven minutes?

As in the previous 2 posts, the first one is catches first train so waits 9..12 minutes or equivalently arrives 0..3 minutes (using the 9 as 3+9=12). The second one is misses the first train so waits 9..19 minutes or equivalently arrives 5..12 minutes (using the 19 as 5+19=24).

How are you getting five when in the mark scheme they got seven?

arrival time is >5 minutes so wait time (before missing first train) is <7.

Why not just consider arrival times of 0, 2, 4, 6, ... and work through what is the wait times for the first and second trains and which intervals are the ones you want? After a few you should spot the pattern.

if he arrives exactly on time he waits 12 minutes, if he arrives 3 minutes late he waits 9 minutes, if he arrives 5 minutes late he waits 7 minutes, and we're considering the probability he waits from 9 to 19 minutes. So the first part is assuming he does arrive 3 minutes late but catches the train, and the second part is assuming he does not catch the first train but waits 7 minutes??

Pretty much as you say,
* first part is when he arrives <= 3 min late and catches first train, so the wait time in 9..12
* second part is arrives >= 5 min late and misses the first train, so wait time is 12..19

Okay, I think I have the principle of it down, the only thing now that I'm confused about is the 5. Sorry if this is a bit repetitive

As in an earlier post, why not put arrival times of 0, 2, 4, .. in and work it thought. But to try and finish it, assuming he arrives 4 minutes late and misses the first train, his wait time is 20 min. Try a few other similar values yourself?

Oh my god, yes I understand it now. If he arrives 5 minutes late he waits 7 minutes and assuming he misses the first one that is, 7 +12, making the wait time a total of 19. Thanks so much!!

Its simple in hindsight, but if youre unsure about stuff like this, stick some values in and work it through.