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

Question

Two persons agree to meet at between 2 PM to 4 PM, but each of them will wait 30 minutes for the late comer. What is the probability that they will meet ?

A. 7/8
B. 1/2
C. 7/16
D. 3/8
E. 7/32

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chetan2u · Answer

We can divide the time period into two parts:-

(1)  Dark Red from 2:30 to 3:30  - If A as shown by indigo lines arrives in this period, then B has full 1 hour to come and meet A as shown within the red lines.
Probability of A is 2:30 to 3:30 in 2:00 to 4:00, so 1 hr in 2hr =  \frac{1}{2} 
Probability of B is also 1 hr in 2hr =  \frac{1}{2} 
Overall probability =  \frac{1}{2}*\frac{1}{2}=\frac{1}{4}

(2)  Turquoise/sky blue from 2:00 to 2:30 and 3:30 to 4:00  - If A as shown by indigo lines arrives in this period, then B has an average of 45 minutes to come and meet A as shown within the red lines. The time period moves from 30 minutes(when A comes at 2:00 or 4:00) to 60 minutes(when A comes at 2:30 or 3:30), so an average of 45 minutes.
Probability of A is 1 hr in 2hr =  \frac{1}{2} 
Probability of B is 45 minutes in 120 minutes =  \frac{45}{120}=\frac{3}{8} 
Overall probability =  \frac{1}{2}*\frac{3}{8}=\frac{3}{16}

Final probability =  \frac{1}{4}+\frac{3}{16}=\frac{7}{16}

C

rocky620 · Answer

Hi  chetan2u 
I approached it as

Case-1   - Lets say A arrives at anytime between 2:00 - 3:30. With each passing minute B will have 30 minutes to come and they will meet.
So, probability =  \frac{1.5}{2}  =  \frac{3}{4}

Case-2   - Lets say A arrives after 3:30. With each passing minute  x , B will have (30-x) minutes to come and they will meet.
So, probability =  \frac{30-x}{30} 
But the value will depend on x.

I am confused how to proceed further in Case-2

Kindly help.

unraveled · Answer

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.

Gknight5603 · Answer

Please find the solution in picture

Posted from my mobile device

KarishmaB · Answer

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)

KarishmaB · Answer

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.

bumpbot · Answer

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

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