A and B are friends. They decide to meet between 1 PM and 2 PM on a given day. There is a
condition that whoever arrives first will not wait for the other for more than 15 minutes. The
probability that they will meet on that day is

• P(meeting)= 15/60 = 1/4
  
  P(not meeting)= 1- 1/4= 3/4
  
  total probability of not meeting= (3/4)*(3/4)= 9/16
  
  so, probability that they meet= 1-(9/16)= 7/16
• 9 years agoHelpfull: Yes(12) No(0)
• wat i hv noticed here is that both the solutions that are given over here are actualy copied from some where.....did any one actualy understood it
• 9 years agoHelpfull: Yes(3) No(0)
• Case-I: A arrives first
  If A arrives in the interval 12:00 to 12:45, then B must arrive within 15 minutes of A’s arrival. The probability of meeting can be found out by considering double definite integral, with limits (0 to 45) and (x to x+15). The probability density for each A and B is 1/60 and 1/60 respectively. By considering these probability densities and finding out the integral under the above-mentioned limits, we get the probability in this part as 3/16.
  If A arrives in the interval 12:45 to 13:00, then B must arrive before 13:00. The probability of meeting can be found out by considering double definite integral, with limits (45 to 60) and (x to 60). By considering the probability densities (1/60 and 1/60 for each A and B) and finding out the integral under the above-mentioned limits, we get the probability in this part as 1/32.
  Summing up the two parts, the probability in the Case-I = 3/16 + 1/32 = 7/32
  Case-II: B arrives first
  Going by the same logic as mentioned in the Case-I, we get
  Probability in the Case-II = 7/32
  Summing up the two cases, the probability = 7/32 + 7/32 = 7/16
• 10 years agoHelpfull: Yes(2) No(1)
• Ans : 7/16
  Let A arrive at x mins past 1
  Let B arrive at y mins past 1.
  
  Now x and y are independent of each other.
  So we need to find P(|x-y|
• 10 years agoHelpfull: Yes(0) No(1)
• can u plz explain in other easy method ?
• 9 years agoHelpfull: Yes(0) No(0)