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Two persons agree to meet at between 2 PM to 4 PM, but each [#permalink]
19 Apr 2006, 14:42

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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 ?

Re: Prob - Two persons [#permalink]
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. _________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

Re: Prob - Two persons [#permalink]
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.

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).

GmatAcer!
"I will either find a way, or make one!"

Re: Prob - Two persons [#permalink]
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. _________________

Who says elephants can't dance?

Last edited by kapslock on 20 Apr 2006, 09:20, edited 1 time in total.

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 ) _________________

Re: Prob - Two persons [#permalink]
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.
for kapslock.

gmatclubot

Re: Prob - Two persons
[#permalink]
20 Apr 2006, 12:38