A and B decide to meet between 5 pm and 7 pm. It was decided that each will wait for the other for 20 mins and if the other person didnt come, he would leave. Find probability of their meeting.

• total time allotted = 2*60 = 120 minutes.
  time during which they can meet= 20 minutes
  so, the probability= 20/120 = 1/6
  
• 10 years agoHelpfull: Yes(45) No(9)
• Incorrect answer MAYANK SHARM,
  This is an example of continuous function as there is no gap between time intervals.
  Consider a graph with A on X axis and B on Y axis with origin at 5 pm
  The end points of each A and B is 2 hrs.
  The possibility of their meeting is given by the area between both of them
  Total area=(2 hrs)^2=120^2
  Region of both A and B is a triangle of base and height equal to 120-20=100
  Reqd. probability=(120^2 - 100^2)/14400 = 11/36 is the required answer..
• 10 years agoHelpfull: Yes(19) No(13)
• hii shruti this is abhishek...tmhari aur meri soch bht milti hai(ans same hai na ummmmmm)...meet me at ..thapa momos.....
• 10 years agoHelpfull: Yes(13) No(25)
• Meeting occurs if the first person arrives between 5:00 and 6:40 and the second
  person arrives in the next 20 minutes after the 1st friend arrives or if both the persons arrive between 6:40 and 7:00.
  Case 1:
  80/120 are the 1st probability
  and hence probability of
  first person arriving
  between 5:00 and 6:40 is
  5/6.
  Probability of second person arriving in the next 20 minutes = 20/120 = 1/6
  person arriving between
  5:00 and 6:40 and meeting
  the other = 5/6 * 1/6*2 =
  5/18( 2 for choosing the first
  arriving friend any one of the two)
  Case 2:
  Both friend
  between 6:40and 7:00. So,
  probability = 1/6*1/6=1/36
  So, probability of a meet = 5/18
  + 1/36 = 11/36
• 8 years agoHelpfull: Yes(3) No(0)
• 1/3
  probability is 20+20/120
• 10 years agoHelpfull: Yes(2) No(1)
• 11/ 36 is the correct answer
  
• 10 years agoHelpfull: Yes(1) No(0)
• This can be solved exactly with the same concept of 'Minimum Length Problem'.
  
  The time period in which they can arrive is 1 hour and they have to be within 20 minutes of each other or 0.333 units of each other.
  
  => The distance from the line A = B should be less than 0.25 units
  
  Required probability = 1 – Unfavorable probability
  
  = 1 – (Sum of the areas of the two triangles / Total Area)
  
  = 1 – 2*(1/2 * 2/3 * 2/3) = 5/9
  
  Please let me know if i am right
• 10 years agoHelpfull: Yes(0) No(7)
• if we want to solve this problem, we calculate for every minute.
  It is impossible
  they are just confusing..
  To this qstn just think in a simple way.
  So,probability of their meeting is just 1/2
• 10 years agoHelpfull: Yes(0) No(1)
• 4*(1/6)*(1/3)+2*(1/6)*(1/6)=1/2
• 10 years agoHelpfull: Yes(0) No(1)
• considering probability curve the max chance of meeting will fall in between the coordinate (100,120) & (120,100) because 120-20=100 min will fall on common region
  
  so the required probability will be (100*100)/(120*120)
• 10 years agoHelpfull: Yes(0) No(0)
• 1-prob of not meet
  =1-((40/60)*(40*60))
  =1-(16/36)
  5/9
  ans 5/9
• 10 years agoHelpfull: Yes(0) No(2)
• b+b=5,tfghghh
  ggyhh
  hhhjjhj
  so b=5
• 10 years agoHelpfull: Yes(0) No(0)
• 0.3298 using graph
• 8 years agoHelpfull: Yes(0) No(0)