Two persons agree to meet at between 2 PM to 4 PM, but each of them wi

I could not get to the answer. This is what i did later(Assuming the meeting happens exactly at 30 minutes).
The earliest both A and B can meet is 2:30pm when either of A or B arrived at 2:00pm. Let's say A comes first and waits for B to join.
As A can arrive at any minute there are 120(2hr=120min) possibilities for arrivals. However, for last 30minutes(3:30pm to 4:00pm) A can't arrive first, in which case B must have arrived earlier or vice-a-versa.
So, for the time period between 2:30 to 3:30(i.e. 60 minutes), Probability of meeting = \(\frac{60}{120} = \frac{1}{2}\) for both.
Hence Probability of meeting = \(\frac{1}{2}*\frac{1}{2} = \frac{1}{4}\)

Now, from 2:00pm to 2:30pm there is a restriction for the meeting as neither A nor B can come before 30 minutes. So, for the time period 3:30 to 4:30, either of them can come earlier than other, starting at 3:01 to 3:30.

Suppose A comes first, Probability of A meeting B is \(\frac{30}{120} = \frac{1}{4}\)
under the condition that B arrives after A from 3:01 to 3:30pm.
Probability of B meeting A = 30/120 = \(\frac{1}{4}\)
Probability of both A and B meeting = \(\frac{1}{4} * \frac{1}{4} = \frac{1}{16}\)

Required probability = \(\frac{1}{4} + \frac{1}{16} = \frac{5}{16}\)
Looks like i did something wrong.

chetan2u I have gone through your solution but don't understand the latter part. Please help.
Bunuel Please do provide official solution.

Thanks.

Please find the solution in picture

Posted from my mobile device

Not a GMAT type question but here is how you can understand it.

A will come at sometime between 2 to 4 i.e. in these 120 mins. Consider them as discrete intervals for ease.

If A comes at 2, B needs to arrive in the next 30 mins so his probability of meeting is 1/4 (because B can arrive at any time in the next 120 mins)
If A comes at 2:01, B needs to arrive either at 2 or in next 30 mins so his probability of meeting is a bit higher than 1/4.
and the probability of meeting keeps increasing till
A arrives at 2:30. B could have arrived at any time from 2 to 3 and they would have met. So probability of meeting is 1/2 now.
This stays the same till A arrives at 3:30 and B can arrive at anytime from 3 to 4.
When A arrives at 3:31, B's probability starts reducing again till it reaches 1/4 when A arrives at 4. B could have arrived only in the interval 3:30 to 4 to meet.

So probability goes from 1/4 to 1/2, stays at 1/2 for half the time and then goes from 1/2 to 1/4 again.

Avg of 1/4 and 1/2 = 3/8
Since probability changes consistently from 2 to 2:30 and from 3:30 to 4, we can say that during these times, it is 3/8 on average.
So for half the time, probability is 1/2 and for other half, it is 3/8.

Total probability = (1/2)*(1/2) + (1/2)*(3/8) = 7/16

Answer (C)

Couple of issues:
- You considered A arrived first but what about cases when B arrived first.
- You will need to average out the probability in the 3:30 to 4 interval above. If you average it out in the 2 to 2:30 interval too, you can consider both cases (A arriving first or B arriving first simultaneously).

Check the solution given above.