Question 65252: Two friends who have unpredictable lunch hours agree to meet for lunch at their favourite restaurant whenever possible. Neither wishes o eat alone and each dislikes waiting for the other, so they agree that each will arrive at a random time between noon and 1pm, and each will wait for the other for 15 minutes or until 1:00. What is the probably that the friends will meet for lunch on a given day?
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! Two friends who have unpredictable lunch hours agree to meet for lunch at their favourite restaurant whenever possible. Neither wishes o eat alone and each dislikes waiting for the other, so they agree that each will arrive at a random time between noon and 1pm, and each will wait for the other for 15 minutes or until 1:00. What is the probably that the friends will meet for lunch on a given day?
You may edit the question.
LET THE 2 FRIENDS BE A AND B.
LET US TAKE OUR UNIT OF MEASUREMENT OF TIME TO BE 1 SEC(THIS COULD BE ANY CONVENIENT INTERVAL). ACCURACY.SO LETUS SPLIT THE TIME INTERVAL OF 12=00 NOON TO 1=00 PM INTO 3600 SECS.THAT IS 0 REPRESENTING 12=00 NOON AND 3600 REPRESENTING 1=00 PM.
LET A COME AT ANY TIME T BETWEEN 0 TO 3600.....TOTAL POSSIBILITIES =3600
LET US CONIDER 3 CASES
1.T IS BETWEEN 0 TO 15*60=900 SECS[12=00 NOON TO 12=15 PM]......900 POSSIBILITIES
2.T IS BETWEEN 900 TO 2700 SECS [12=15 PM TO 12=45 PM.].......1800 POSSIBILITIES
3.T IS BETWEEN 2700 TO 3600 SECS [12=45 NOON TO 13=00].........900 POSSIBILITIES
CASE1…
B CAN COME FROM 0 TO T+900 ANY TIME….NUMBER OF POSSIBILITIES
= (T+900)-0 = T+900
CASE 2
B CAN COME FROM T-900 TO T+900 ANY TIME ..NUMBER OF POSSIBILITIES
=(T+900)-(T-900)=1800
CASE 3
B CAN COME FROM T-900 TO 3600 ANY TIME..NUMBER OF POSSIBILITIES
=3600-(T-900)= 2700-T
TOTAL POSSIBILITIES = 3600*3600 = 12960000
SUCCESFULL POSSIBILITIES..= 900(T+900)+1800*1800+900(2700-T)= 6480000
PROBABILITY OF SUCCESS OR HAVING LUNCH TOGETHER = 6480000/12960000 = 0.5
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