Thread: Please check if my answer is correct..

Hi everyone.. I solved this problem but i'm not too confident..hope somebody can check and if it's wrong, please help me solve it the right way..

Afra and Wiro takes the train to their 8:30 AM class and arrives at the station uniformly between 7:00 AM and 7:20 AM. They both agreed that they are willing to wait for one another for 5 minutes, after which they take the train and ride alone or together. Assume that they always arrive randomly during that specified period and they don't communicate whatsoever before meeting. What is the probability that they will meet and ride the train together?

*Hint given by our professor: In a cartesian plane, a square of side 20 (minutes) represents all the possibilities of the morning arrivals of Wiro and Afra at the train station. For example (01, 07) means they arrive at 7:01 and 7:07, hence they ride the train alone. Now, the probability that they will meet= Area of region of meeting divided by the area of the square.

My answer is:
There are 420 possible morning arrivals and out of those 420 possibilities, there are 105 possibilities that they arrive within 5 minutes between each other. Therefore, i concluded that the probability that they will meet each other is 0.25.

*Based on my professor's hint, i also assumed that the area of the square=420 while the area of the region of the meeting=105. Hence, 105 divided by 420 is 0.25.

Please correct me if i'm wrong..please..thanks..

My answer is:
There are 420 possible morning arrivals and out of those 420 possibilities, there are 105 possibilities that they arrive within 5 minutes between each other.

How did you decide there were 105 possibilities?


If Afra arrives at 7:00 then Wiro has a 6 possibilities (7:00, 7:01, 7:02, 7:03, 7:04, 7:05) of arriving within Afra's waiting time.

If Afra arrives at 7:01 then Wiro has a 7 possibilities (7:00, 7:01, 7:02, 7:03, 7:04, 7:05, 7:06) of arriving within Afra's waiting time.


It seems to me that the closer Afra arrives to the 7:05-7:15 range, the better the chances get.

You could consider making a table of the 21 cases (don't panic, the 11 cases in the middle seem to have the same outcome so it's not a ton of work).

I haven't done the work to calculate this but I believe they have better than 25% odds for riding together.

Hey wytiaz.. sorry i think i did it wrong.. there should be 441 possibilities .. and there can only be 6 possibilities that they will meet from each case.. so 6 multiplied by 21 cases= 126. so i think it's 126/441 = 0.2857143..what do you think?

Look at a sample case.

Afra arrives at 7:13

Which times could Wiro arrive allowing them to ride together?

7:08 (Wire has been waiting for Afra)
7:09 (Wire has been waiting for Afra)
7:10 (Wire has been waiting for Afra)
7:11 (Wire has been waiting for Afra)
7:12 (Wire has been waiting for Afra)

7:13 (they arrive together)

7:14 (Afra waits for Wiro)
7:15 (Afra waits for Wiro)
7:16 (Afra waits for Wiro)
7:17 (Afra waits for Wiro)
7:18 (Afra waits for Wiro)


There are 21 discrete times that Afra could arrive. Investigate them and see if you can come up with a table that shows how many of Wiro's times could let them go to school together. Then sum the possibilities and divide by 441.

I did a list of the possible times that afra arrives and waits for wiro so they can ride together. I got 111 possibilities that they can ride together if afra will wait for wiro or if wiro will arrive at the same time as afra..

What should I do now? Multiply that number by two since if i list down the possibilities that they will ride together if wiro will wait for afra, i will also get 111 possibilities?

If I use 111 as the # of possibilities that they will ride together, I will still come up with 25%..

However if I use 222, it'll be 50%..

What do you think?