A bus arrives every 20 min to bus stop:beginning at 6.40am and continuing till 8.40am. A passenger, not knowing the schedule, randomly arrives between 7am and 7.30am. what is the probability that he waits for more than 5 min for bus?

• 
  Let the bus arrives at 7:20 then the possibe timing in which passenger can reach the stand are 7:01..7:02...........7:20...total=20 .
  Out of this the passenger have to wet more then 5 mint if he raach there 7:01-7:14...possible outcomes 14.....
  If bua arrives at 7:20 the ...p=14/20
  If arrives at 7:40 then possible probability would be more then 1....confusing
  
• 11 years agoHelpfull: Yes(0) No(0)
• Wet ...probability off his arrivals at bus stand at of before 7:20 and between 7:21-7=30 are
  2/3 and 1/3 respectively
  Now if he arrives between 7:01-7:21 ,probability of his waiting more then 5 mint would be 2/3*14/20=28/60
  Again if he catch the bus at 7:40 then 1*1/3
  So required probability =28/60+1/3=48/60=4/5 ans
• 11 years agoHelpfull: Yes(0) No(0)
• he waits more than 5 mins if he reached.......
  7:1,7:2,7:3,7:4,7:5,7:7,7:8,7:9,7:10,7:11,7:12,7:13,7:14,
  so prob=1/14 ans...
• 11 years agoHelpfull: Yes(0) No(0)
• he waits more than 5 mins if he reached.......
  waits more than 5 mins...
  7:1,7:2,7:3,7:4,7:5,7:7,7:8,7:9,7:10,7:11,7:12,7:13,7:14,
  n after 7:21,7:22,7:23,7:24
  so prob=1/18 ans
  
• 11 years agoHelpfull: Yes(0) No(0)
• he waits more than 5 mins if he reached.......
  waits more than 5 mins...
  7:1,7:2,7:3,7:4,7:5,7:7,7:8,7:9,7:10,7:11,7:12,7:13,7:14,
  n after 7:21,7:22,7:23,7:24
  so prob=18/29
• 11 years agoHelpfull: Yes(0) No(0)
• ANS:-19/31(7:00,7:01,7:02,7:03,7:04,7:11,7:12,7:13,7:14,7:15,7:16,7:18,7:19,7:20,
  7:21,7:22,7:23,7:24 ALL THIS ARE COUNTED AS WAIT FOR MORE THAN 5 MINS)
  BUS ARRIVES AT 7:10 AND 7:30
• 11 years agoHelpfull: Yes(0) No(0)
• Bus only can arrive at an interval of 20 mint ...
• 11 years agoHelpfull: Yes(0) No(0)