Independent events in a real life scenario The question:
Martin's journey home from work involves taking the bus to the station, catching the six o'clock train and then a short walk to his house. If the bus is late, he has to catch a later train.
On 10 days out of the past 50, the bus has not reached the station in time for him to catch the six o'clock train. Similarly on 10 days out of the past 50, the six o'clock train has arrived late at Martin's home station.
Is it more likely than not that Martin will get home on time tomorrow?
A. Yes, the buses and the trains each meet the requirements of
the timetable 40 out of 50 times respectively, so he should get home
on time tomorrow.
B. Yes, if the bus is late, the train may be too, so he would get home on time.
C. No, on 20 occasions over the 50 days either the train or bus has been late, so it is unlikely he will get home on time.
D. No, every day is different so predictions are no more than guesses and wholly unreliable.
In my opinion, B and C are obviously incorrect answer options. I chose D, because I thought whether or not the train / bus comes late tomorrow, is independent of past probabilities. But obviously, if you use the data given in the information, A is the correct answer. $ \frac{40}{50} \times \frac{40}{50} = 0.64 $
Could someone explain why the answer is A?
Like, let's say there are 4 white balls and 4 black balls in a bag. If you take one ball out and it was white, you put it back in the bag, and if you take one out again, the probability that the ball is white is still $\frac{1}{2}$. (Independent events).
Is this scenario different to the balls in a bag scenario?
 A: Letting $A$ be the event the bus is late and $B$ the event the train is late.  Martin being late corresponds (presumably) to the event $A\cup B$, the bus or train being late.  Assuming what he has experienced over the past 50 days is indicative of what he will likely experience in the future as well, we know that $\Pr(A\cup B) = \Pr(A)+\Pr(B)-\Pr(A\cap B)\leq \Pr(A)+\Pr(B)=0.2+0.2=0.4$ so at worst we know he is late at most $40\%$ of the time.  Note that this did not use anything about independence.
If it were true that these were independent, then we could say further that $\Pr(A\cup B)=\Pr(A)+\Pr(B)-\Pr(A\cap B)=\Pr(A)+\Pr(B)-\Pr(A)\Pr(B)=0.2+0.2-0.2\cdot 0.2 = 0.36$
It is not a good assumption in real life however to assume that these are independent events as there may be reasons that cause both buses and trains to be late for the same reason, for instance if there were a major sporting event occurring that clogs traffic of all kinds, or some disastrous weather.
Arguably, it might not be a safe assumption that the past 50 days are going to be indicative of what he is likely to experience in the future either.  For instance, if there was major road construction or multiple sporting events over the past 50 days which can be pointed to as the cause of the delays to bus and train schedules which might no longer be a factor moving forward.
If we were to make the assumption that the past 50 days are indicative of what he'll experience moving forward I would side with (A) as the correct answer as even if we were to ignore the question of independence, we find he will be late at most 40% of the time and on time at least 60% of the time and so will be more likely than not to be on time.
If we were not to make the assumption that the past 50 days are indicative of what he'll experience moving forward, I would side with (D) as the correct answer as we are not given any information as to the actual probability distributions involved.
A: We have $50$ samples for each event, so the quoted event counts can reasonably be taken as representatives of the overall punctuality of each service and a probability extracted. Furthermore, for the purposes of this question bus and train may be safely assumed as independent (you can't drive a bus and train at the same time, the two services run on different infrastructure and do not influence each other, etc.) so we have independent random variables and may multiply to arrive at $0.64$.

I chose D, because I thought whether or not the train / bus comes late tomorrow, is independent of past probabilities.

To choose D is to imagine that the bus or train drivers conspire to prevent Martin from reaching home on time, which violates Occam's razor.
A: I would do this: imagine 5000 such scenarios.  "10 out of every 50" is 1/5 so in 1000 such situations, the bus arrives at the station late, in 4000, the bus is on time.  In 1/5 of the times the bus is late, so 200, the train is also late so he can take it anyway.  There are 4000+ 200= 4200 times out of 5000 that he catches the train.  The probability he will get home on the first train is 4200/5000= 42/50= 84/100 or 84%.
