Question

Alice and Bob arrange to meet for lunch on a certain day at noon. However,neither is known for punctuality. They both arrive independently at uniformly distributedtimes between noon and 1 pm on that day. Each is willing to wait up to 15 minutes forthe other to show up. What is the probability they will meet for lunch that day? "slader"

Answer 1

Answer:

probability of there meeting = 15/(60-15) = 15/45 = 0.333 = 33%

Step-by-step explanation:

favorable event = +- 15 minutes

total time they can eat launch is between noon and 1pm = 60minutes

they are both known for non punctuality, that means they cant meet up at 12noon, let assume 15minutes pass noon.

probability of there meeting = 15/(60-15) = 15/45 = 0.333 = 33%

which is probably saying they will not meet for lunch that day.