Question

Today there are five trains labeled 1 to 5 that come to a train

station in a uniformly random order. Two of your friends come to a train

station, and you have no idea have many trains (of these five), if any, have

already left - any possibility is equally likely. Each of your friends have

the intention of boarding the first train which will take him/her to his/her

destination. The trains which take your first friend to his destination are

those labeled with 1, 2 and 3, and trains that take your second friend to her

destination are those labeled with 2, 3 and 4. If you learn that they both

ended up boarding the same train, what is the probability that they arrived

at the train station just before the first train of the day?

Answer 1

Solution.

Let Ao be the event that they arrived before the first
train
A1 be the event that they arrived after the first train

As be the event that they arrived after the 5th train
IF B is the event that they boarded the same train we
need to find
P(Ao |B)we use the Bayes rule.We know that the
1
probability of any event Ai is 6
Moreover we have P(B|A5)=0
Since if A5 happened there are no more trains left
The probability(B|4)is the probability that the last
2
train is either 2 or 3 which has the probability 5

fourth train was either 2 or 3(which has the probability
2
5)or that the fourth train was 5 and fifth train was(2 or
1
3)which has the probability 10.
12 1
P(B|A3)=10+5-2
Therefore
If they arrive right after the second train then again they
could have boarded The Train Together if third train was
2 or 3,or if 3rd train was 5 Or the 4th one 2 or 3.It is
simple observation that they could not have boarded
the last train because there is only one train(no 5)they
P(B|A2)=5
both want to miss.Therefore
Which is also true for P(B|A1)and P(B|Ao)=5