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Man and woman agree to meet at a certain location at 12:18 pm. if the man arrives at a time that is uniformly distributed between 12:07 pm and 12:29 pm, and if the woman arrives independently at a time that is uniformly distributed between 12:00noon and 1:00 pm, what is the probability that the man arrives first?

X: The arrival time of the man u~(15,45)
and
Y: The arrival time of the woman uni~(0,60)
Can anyone explain to me why I have to calculate the P(Xwhat I want to know is why the event X corresponds to the event that the man arrives first.

User Booger
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1 Answer

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Final answer:

To calculate the probability that the man arrives first, we assess the uniformly distributed arrival times of both individuals. The man's arrival time, X, is uniformly distributed between 15 and 45 minutes past noon, while the woman's arrival time, Y, is uniformly distributed from noon to 1:00 pm (0 to 60 minutes past noon).

Step-by-step explanation:

To determine the probability that the man arrives first, we need to compare the uniformly distributed arrival times of the man and the woman. The man's arrival time, X, is given to be uniformly distributed between 12:07 pm (which is 12:00 pm + 7 minutes, or 12:00 pm + 7/60 hours, thus 7 minutes past noon) and 12:29 pm (12:00 pm + 29 minutes, or 12:00 pm + 29/60 hours, thus 29 minutes past noon). This can be expressed as X being uniformly distributed between 15 minutes and 45 minutes past noon, as there are 15 minutes from noon to 12:15 pm and from there an additional 7 minutes to 12:07 pm, making the time 22 minutes and likewise for 12:29 pm, giving us X~U(15, 45). The woman's arrival time, Y, is uniformly distributed between 12:00 pm (0 minutes past noon) and 1:00 pm (60 minutes past noon), hence Y~U(0, 60).

User Myborobudur
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