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2 August, 16:14

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"

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  1. 2 August, 18:01
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    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.
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