Probability of meeting

mr.wong · 2016-09-14 15:02:42

A boy and a girl have dated to meet at a place .
They will arrive there randomly within 60 minutes
and are willing to wait for one another for 20 mins.
What is the probability of their meeting at the place?

( Solution illustrated with a diagram will be appreciated .)

bobbym · 2016-09-14 17:08:44

Hi;

mr.wong · 2016-09-14 23:01:19

Thanks bobbym ,

Your answer is correct !
Now let us ask another question : What will be the expectation of the
waiting time of any one of them , say the boy ?

Your diagram may be divided into 3 portions . If the blue portion
is bisected by the diagonal ( the curve y = x ) then it becomes 4
portions .
Let A , B , C and D denote the 4 portions one by one from the
south - east triangle to the north - west one .
Let the boy arrives at x minutes and the girl arrives at y minutes .
For A , x - y ≥ 20 . The boy does not see the girl , but still expecting she will come latter . He will wait for min ( 20 , 60-x ) minutes .
For B , x - y ≤ 20 , the boy need not wait , i.e. he will wait for 0 minute .
For C , y - x ≤ 20 , the boy will wait for y-x minutes .
For D , y - x ≥ 20 , the boy will keep waiting for 20 minutes .

How should we continue ?

bobbym · 2016-09-15 06:20:31

Hi;

A boy and a girl have dated to meet at a place .
They will arrive there randomly within 60 minutes
and are willing to wait for one another for 20 mins.
What is the probability of their meeting at the place?

Supposing the boy has to wait 28 minutes? What do you want to do with that possibility? Would the boy have left after 20 minutes? That would make the maximum waiting time for him of 20 minutes.

mr.wong · 2016-09-15 14:56:08

Hi bobbym ,

You are right . The maximum waiting time for the boy is 20 mins .
He will left immediately after 20 mins . even he knows that the girl will
come 8 mins. later .

bobbym · 2016-09-15 15:15:19

Also, when the girl gets there and waits 20 minutes she will leave, now the boy comes, he does not know she was there so he waits 20 minutes also and then leaves. Is that how you want to handle that situation?

mr.wong · 2016-09-16 15:13:00

Hi bobbym ,

Yes , that's the case .

mr.wong · 2016-09-16 15:17:41

To find the expectation of the boy's waiting time , we may use a
3-dimensional diagram with base of bobbym's diagram and the
height at z-axis showing the corresponding waiting time .

(1) For portion ( A ) , it may be re-divided into 3 regions .
(i) region A1 denotes the triangle (20,0) , (40,0) and (40,20)
with height being 20 ( minutes) and area = 20 * 20 * 1/2 = 200 .
Thus its volume = 20 * 200 = 4000 .

(ii) region A2 denotes the square (40,0) , (60,0) , (60,20) and (40,20) with heights being 20 at one side and 0 at the opposite
side . Thus the volume of this prism = 20* 20 * 1/2 * 20 = 4000 .

(iii) region A3 denotes the triangle (40,20), (60,20) and (60,40) .
with the corresponding pyramid with height being 20 .
Thus the volume = 1/3 * 20 * 20 * 20 * 1/2 = 4000 / 3 .
Thus total volume over A will be 4000 + 4000 + (4000 / 3 )
= 8000 + (4000 / 3 )

(2) For portion (B) , since the corresponding height = 0 ,
therefore the corresponding volume also = 0 .

(3) For portion (C) , it may also be divided into 3 regions .
(i) region C1 denotes the rectangle with length ( 0,20) and (40,60) ,
= √2 * 40 . and width = √2 * 10 .
Volume of corresponding prism = 20 * √2 * 10 * 1/2 * √2 * 40
= 8000

(ii) region C2 being a triangle at one side of the rectangle with area
√2 * 10 * √2 * 10 * 1/2 = 100 . Thus volume of the corresponding pyramid = 1/3 * 20 * 100 = 2000 / 3

(iii) region C3 being another symmetric triangle at the other side of
the rectangle . Its area and volume of the corresponding pyramid are
identical to those of region C2 . Thus the corresponding volume is
also 2000 / 3 .
Thus total volume over C will be 8000 + ( 2000 / 3) * 2
= 8000 + (4000 / 3 ) ( same as A .)

(4) For portion D , its area = 40 * 40 * 1/2 = 800 . Height of the
corresponding prism = 20 , thus its volume = 800 * 20 = 16000 .

Thus the total volume over the base = 16000 + 8000 / 3 + 16000
= 32000 + 8000 / 3 = 104000 / 3

So the expectation of the waiting time of the boy
= 104000 / ( 3 * 60 * 60 )
= 260 / 27 mins .
= 9 + 17/ 27 mins . ( about 9.63 mins . )

bobbym · 2016-09-16 15:23:09

Hi;

I am not getting that.

mr.wong · 2016-09-16 21:39:41

Hi bobbym ,

What result did you get ?
My result may be incorrect for I have not checked it carefully ,
but the procedure should be ok .

mr.wong · 2016-09-16 23:16:15

Hi bobbym ,

I have found where I made a mistake !
The total volume over the base should be
16000 + 8000 + ( 4000 / 3 ) + 16000
= 40000 + ( 4000 / 3 )
= 124000 / 3

So the expectation of the waiting time of the boy
= 124000 / ( 3 * 60 * 60 )
= 310 / 27 mins .
= 11 + 13 / 27 mins. ( about 11.5 mins )

Hope this answer be correct .

bobbym · 2016-09-17 06:18:48

That is correct! But I did it in another way.

mr.wong · 2016-09-17 16:52:49

Hi bobbym ,

I am confused now ! It seems my answer in # 8 should be correct !
( I had not made any mistake . ) In fact I made a mistake in # 11 .
Please check your work .

bobbym · 2016-09-17 16:56:58

The answer in post #11 agrees with my simulation and my analytical solution. I believe it is the correct answer.

thickhead · 2016-09-17 20:59:58

I get 5 minutes average waiting time.

mr.wong · 2016-09-18 15:53:06

Hi bobbym ,

I have checked my solution in # 8 for several times but I can't
find where I made a mistake .

mr.wong · 2016-09-18 15:54:10

Hi thickhead ,

Please show your procedure .

bobbym · 2016-09-18 19:06:43

You do not know what the answer is?

mr.wong · 2016-09-18 22:05:23

Hi bobbym ,

No , I don't .

bobbym · 2016-09-19 00:15:02

Hmmm, that is a problem.

thickhead · 2016-09-19 18:22:35

Simple approach:-
From boy's point of view.His arrival time is t.
(a)t=20 to t=40 minutes. the girl may show up at any time between t-20 to t+20. for half this time his waiting time is 0. For the other half it varies from 0 to 20 averaging 10 minutes. For the 2 periods combined the average is 5 minutes.
(b)At t=0 he has to wait for anything from 0 to 20 m.average waiting time=10 m. From t= 0 to t=20 we can assume linear variation from 10 to 5 minutes.
(c)At t=60 waiting time =0. assume linear variation from 5m at t=20 t0 0 at t=60
the average as whole={(10+5)/2*20+5*20+(5+0)/2*20}/60= 5 minutes.

Last edited by thickhead (2016-09-20 18:25:57)

mr.wong · 2016-09-20 17:49:24

Hi thickhead ,

I wonder whether the problem can be solved in your way .
You should take reference to bobbym's diagram . The weight
( area ) of various regions are not fixed thus you can't just take
simple average .
E.g. in (a) from t = 20 to t = 40 , only 1/3 but not half the time
= 0 . Another 1/3 the average time = 20
while the remaining 1/3 the average time = 10 ( ? ) .
Thus the combined average in (a) should 0 + 20 + 10 ) / 3 = 10 .

thickhead · 2016-09-20 17:58:31

Last edited by thickhead (2016-09-20 17:59:36)

thickhead · 2016-09-20 18:30:03

mr.wong wrote:

Hi thickhead ,
I wonder whether the problem can be solved in your way .
You should take reference to bobbym's diagram . The weight
( area ) of various regions are not fixed thus you can't just take
simple average .
E.g. in (a) from t = 20 to t = 40 , only 1/3 but not half the time
= 0 . Another 1/3 the average time = 20
while the remaining 1/3 the average time = 10 ( ? ) .
Thus the combined average in (a) should 0 + 20 + 10 ) / 3 = 10 .

Half the time accounts for this. If the girl comes earlier by 0 to 20 minutes his waiting time is 0.If she comes later between 0 to 20 minutes the average waiting time is 10. the average of these two is 5.

bobbym · 2016-09-20 20:01:17

What about the times the boy must wait 20 minutes. That adds to the average you get of 5 minutes. Does this not suggest that your answer of 5 or less is not correct?

Math Is Fun Forum

#1 2016-09-14 15:02:42

Probability of meeting

#2 2016-09-14 17:08:44

Re: Probability of meeting

#3 2016-09-14 23:01:19

Re: Probability of meeting

#4 2016-09-15 06:20:31

Re: Probability of meeting

#5 2016-09-15 14:56:08

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#6 2016-09-15 15:15:19

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#7 2016-09-16 15:13:00

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#8 2016-09-16 15:17:41

Re: Probability of meeting

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#10 2016-09-16 21:39:41

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#11 2016-09-16 23:16:15

Re: Probability of meeting

#12 2016-09-17 06:18:48

Re: Probability of meeting

#13 2016-09-17 16:52:49

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#14 2016-09-17 16:56:58

Re: Probability of meeting

#15 2016-09-17 20:59:58

Re: Probability of meeting

#16 2016-09-18 15:53:06

Re: Probability of meeting

#17 2016-09-18 15:54:10

Re: Probability of meeting

#18 2016-09-18 19:06:43

Re: Probability of meeting

#19 2016-09-18 22:05:23

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#20 2016-09-19 00:15:02

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#21 2016-09-19 18:22:35

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#22 2016-09-20 17:49:24

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#23 2016-09-20 17:58:31

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