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

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Two persons agree to meet at between 2 PM to 4 PM, but each [#permalink] New post 19 Apr 2006, 14:42
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 ?
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Re: Prob - Two persons [#permalink] New post 19 Apr 2006, 15:26
Suppose the guys are A and B.

Probability A comes first = 1/2.
If A comes first and waits 30 mins, he waits for 1/4th of entire meeting time, and he meets B if B's 30 mins overlap these 30 out of 120 mins. Thus the time fraction in which A and B meet = 60/120.

Similarly for the case of B coming first.

Thus probability = 1/2*1/2 + 1/2*1/2 = 1/2.
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 [#permalink] New post 19 Apr 2006, 16:28
Agree with kapslock.
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Re: Prob - Two persons [#permalink] New post 19 Apr 2006, 19:25
kapslock wrote:
Suppose the guys are A and B.

Probability A comes first = 1/2.
If A comes first and waits 30 mins, he waits for 1/4th of entire meeting time, and he meets B if B's 30 mins overlap these 30 out of 120 mins. Thus the time fraction in which A and B meet = 60/120.

Similarly for the case of B coming first.

Thus probability = 1/2*1/2 + 1/2*1/2 = 1/2.


The official answer is not 1/2.

Also, what happens if A comes after 3:30?
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 [#permalink] New post 20 Apr 2006, 04:33
Consider the two persons are X and Y.

Probability that X or Y comes first = 1/2

Probaility of X coming first and waiting for 30 min= 1/2 (30/120) = 1/8

Probaility of Y coming first and waiting for 30 min= 1/2 (30/120) = 1/8

Probaility of meeting X and Y = 1/8 + 1/8 = 1/4
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 [#permalink] New post 20 Apr 2006, 05:02
Another vote for 1/4 , but different explanation.

number of possibilities they meet: 120*30 ( first come any time and then second come in 30 min. )

total : 120 * 120

probability: meet/total = 120 * 30 / (120 * 120) = 1/4
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 [#permalink] New post 20 Apr 2006, 05:24
Can we realy apply probability principles to this kind of problems? I am not able to find the approach.

Give that the time is between 2:00 PM and 4:30 PM and assuming that both people respect each other's time, probability of one person meeting another is 1/2 (if other person is within other's 30 mins wait time, they will meet and if other person is not within other's wait time he will not meet. So we have two possible outcomes).

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 [#permalink] New post 20 Apr 2006, 05:29
I think it shud be 1/2.


The probability that A will come within a 30 minute period is 1/4
Then B is left with no choice but to come in that 30 min period to meet A

So its 1/4*1 = 1/4

Now reversing the order if B comes before A, we again get 1/4

So 1/4+ 1/4 = 1/2
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 [#permalink] New post 20 Apr 2006, 06:05
I think the right answer is 7/16. You can't assume here that they come
in intervals of one minute. So the approach should be different.
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Re: Prob - Two persons [#permalink] New post 20 Apr 2006, 07:43
giddi77 wrote:
vipin7um wrote:
kapslock wrote:
Suppose the guys are A and B.

Probability A comes first = 1/2.
If A comes first and waits 30 mins, he waits for 1/4th of entire meeting time, and he meets B if B's 30 mins overlap these 30 out of 120 mins. Thus the time fraction in which A and B meet = 60/120.

Similarly for the case of B coming first.

Thus probability = 1/2*1/2 + 1/2*1/2 = 1/2.


The official answer is not 1/2.

Also, what happens if A comes after 3:30?


Is it 3/8?


Hmm giddi, I too think our answers didn't consider the fact that A could arrive after 3:30.Here's new calculation:

Considering case of A coming first, and waiting for 30 minutes to meet the other person. However, for the last half hour 3:30pm to 4:00pm, the probability of meeting B fades because B would wait till 4:00 and no more.

Considering the area of the graph would represent the probability (the graph on a constant height of 1 from 2:00 to 3:30 and then linearly declining to 0 at 4:00). This area is 6/8 + 1/2*2*1/8 = 7/8.

Therefore, for A, the probability is (1/2)*(60/120)*(7/8) = 7/32.
Ditto for B.
Thus total probability = 7/16.
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Last edited by kapslock on 20 Apr 2006, 09:20, edited 1 time in total.
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 [#permalink] New post 20 Apr 2006, 08:02
Let's use minute granularity:

P( difference amongst the arrivals within the 2 hours = exactly 30 min )= 91/( 121*61 )
P( difference amongst the arrivals within the 2 hours > 30 min ) = 45*91/( 121*61 )
P( difference amongst the arrivals within the 2 hours <= 30 min ) = 1 - 45*91/(121*61) = 106*31/ ( 121*61 )
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Re: Prob - Two persons [#permalink] New post 20 Apr 2006, 12:38
kapslock wrote:
giddi77 wrote:
vipin7um wrote:
kapslock wrote:
Suppose the guys are A and B.

Probability A comes first = 1/2.
If A comes first and waits 30 mins, he waits for 1/4th of entire meeting time, and he meets B if B's 30 mins overlap these 30 out of 120 mins. Thus the time fraction in which A and B meet = 60/120.

Similarly for the case of B coming first.

Thus probability = 1/2*1/2 + 1/2*1/2 = 1/2.


The official answer is not 1/2.

Also, what happens if A comes after 3:30?


Is it 3/8?


Hmm giddi, I too think our answers didn't consider the fact that A could arrive after 3:30.Here's new calculation:

Considering case of A coming first, and waiting for 30 minutes to meet the other person. However, for the last half hour 3:30pm to 4:00pm, the probability of meeting B fades because B would wait till 4:00 and no more.

Considering the area of the graph would represent the probability (the graph on a constant height of 1 from 2:00 to 3:30 and then linearly declining to 0 at 4:00). This area is 6/8 + 1/2*2*1/8 = 7/8.

Therefore, for A, the probability is (1/2)*(60/120)*(7/8) = 7/32.
Ditto for B.
Thus total probability = 7/16.


Wonderful explanation. Right answer is 7/16.
:wave for kapslock.
Re: Prob - Two persons   [#permalink] 20 Apr 2006, 12:38
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