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?
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.
I would do this: imagine 5000 such scenarios.  &quot;10 out of every 50&quot; 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%.