I have two servers at a grocery store. P(server 1) is 0.4. P(server 2) is 0.6. P(time is less than 5 min | server 1) is 1. P(time is less than 5 min | server 2) is 4/9. What is P(server 1 | time is less than 5 min)? Would it be 1/(1+ 4/9) or do the weightings from before come into play?
2006-09-13 08:52:32 · 3 answers · asked by kweise 2 in Science & Mathematics ➔ Mathematics
Not all less than 5 minute visits are with server 1, but most are.
The probability of a <5 visit is (0.4)*(1)+(0.6)*(4/9). Notice it would be 1 if both servers were efficient!
We're going to assume (normalize) this to 1.
2006-09-13 09:14:56 · answer #1 · answered by bubsir 4 · 0⤊ 1⤋
I am unclear what you have here, but it does not appear to be a well-defined queuing (waiting line) problem. What are the arrival and service rates, for example? Is this a single line, multiple server or single line, single server problem? Need more scenario information before your problem can be worked.
2006-09-13 09:04:40 · answer #2 · answered by oldprof 7 · 0⤊ 1⤋
you're question doesnt make sense. lay it out in an easier way to read and i'll have another look.
2006-09-13 09:04:58 · answer #3 · answered by Schorpe 2 · 1⤊ 1⤋