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Probability: joint probability distribution problem? |
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| Apr8-10, 10:56 AM | #1 |
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Probability: joint probability distribution problem?
1. The problem statement, all variables and given/known data
John and George are set to meet each other at 12 o'clock. John's time of arrival, J, is distributed uniformly between 12:00 and 12:15. John will wait for George for 15 minutes. If he doesn't show up, he leaves. George's time of arrival, G, is also uniformly distributed, between 12:05 and 12:20. But he will only wait 5 minutes for John. Find the probability of an encounter. 3. The attempt at a solution I'm at complete lost with this problem. I think that what I have to do is find the jount probability distribution of J and G, but I couldn't say why. And the "waits for 15 minutes" thing also confuses me. Any ideas? |
| Apr8-10, 02:30 PM | #2 |
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Anybody?
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| Apr8-10, 03:39 PM | #3 |
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Hint: Let T=G-J. For what range of values of T will an encounter happen?
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| Apr8-10, 05:10 PM | #4 |
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Probability: joint probability distribution problem?
T between 0:00 and 0:05...
OK, and I have to do the same thing but considering T2 = J-G. How can I then find the entire answer? T+T2? |
| Apr8-10, 05:47 PM | #5 |
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Note that T2 = -T, so it's essentially the same variable. In other words, if you have something like a<T2<b, that's the same as -a>T>-b. You only have to work with one variable. There's no need to work the cases out separately and combine them at the end.
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| Apr10-10, 11:26 AM | #6 |
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No, no, I see. T has to be between -0:15 and 0:05, right? Because if G<J, T<0, and since J can only be 15 minutes earlier than G otherwise he leaves, T> -0:15. and if G>J, T>0, and since G can only be 5 minutes earlier than J, T<0:05.
So, what I'm looking for is P(-0:15<= T <= 0:05), right? |
| Apr10-10, 02:31 PM | #7 |
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Right!
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| Apr11-10, 10:21 AM | #8 |
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OK, thanks.
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| Sep13-10, 01:16 AM | #9 |
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All understood.. but how to find out the p.d.f of T? Because without the pdf of T the required probability cannot be calculated..
Please help |
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