00:01
So we are going to consider a key -in system with two servers.
00:05
So you can see that it has an exponential inter -arrival time distribution with mean equal to two hours.
00:17
So therefore, the mean rate of the interval time will equal to lambda n to be 1 over the mean, to the mean arrival time.
00:32
So here will be for n greater than not equal to 0.
00:35
So this would be equal to 1 over 2 for n greater than not equal to 0 so we can see it as an exponential service time distribution with mean of 2 so this implies that the main service rate so we get mean n to be equal to 1 over the mean so the mean service time for n is equal to 1 so let me just bring this and it has 1 over the mean service time plus one over the mean service time for n greater than not equal to two so this implies getting mean of n to be equal to so we can 1 over 2 for n is equal to 1 and we can 1 over 2 plus 1 over 2 for n greater than or equal to 2 so this implies getting 1 over 2 for n is equal to 1 and 1 for n greater than not equal to 2.
01:44
So when we come to the part i, which says that a customer has just arrived at 12 o 'clock.
01:50
Here, the objective is to determine the probability that the next arrival will come before 1 o 'clock.
01:57
So this means that the next arrival will come within 1r.
02:02
So the cumulative probability here will be the probability of t less or equal to small t, which is equal to 1 minus e to the power minus lambda t.
02:12
So we know what lambda is to be 1 over 2 and our t is giving us 1.
02:22
And the cumulative probability formula will also be the probability of t less or equal to t to be equal to 1 minus e to the power minus mu.
02:33
T which represents the next r within t hours.
02:38
So to find the probability of next arrival before 1 p .m.
02:49
Will be equal to 1 minus e to the pound minus 1 over 2 times 1.
02:54
So we get 1 minus 0 .6 .5 which is equal to 0 .3934.
03:02
So here the required probability is 0 .3934.
03:06
When we come to the ii, the objective is to determine the probability that the next arrival will come between 1 o 'clock and 2.
03:16
So we get the probability of the next arrival between 1 o 'clock or 1 p .m.
03:25
2 p .m.
03:27
To be equal to 1 minus the probability of the next arrival before.
03:32
So let me write before 1 p .m.
03:38
Minus the probability of next arrival after 2 after 2 p .m.
03:45
So this would be called 1 minus 0 .393 .4.
03:50
Minus e to the power 1 over 2 times 2.
03:54
So we get 0 .606 minus 0 .3979 and we get 0 .2380 and we get 0 .2380.
04:07
So this is the required probability.
04:09
When we come to the i .i, you have to determine the probability that the next arrival will come after 2.
04:16
So this means that the arrival will come after 2 hours.
04:21
So our cumulative for average is the probability of t will be greater than or equal to t to be equal to e to the power minus lambda t.
04:30
So we know what our lambda in our t is.
04:32
So our probability of the next arrival after 2 p .m.
04:39
Will be equal to e to the power 9 is 1 over 2 times 2.
04:43
So we get 0 .39.
04:47
Sorry, 36, 7, 9.
04:49
So this is our probability.
04:51
When we come to the question two, you have to determine the probability that the next arrival will come between 1pm and 2.
05:02
So here we're going to as you know, other customer will arrive before 1 o 'clock p .m.
05:10
So the probability that the next arrival between 1 p .m.
05:14
And 2 p .m.
05:15
Given lambda will be.
05:16
So let me write the probability that our next arrival will be between 1 p .m.
05:24
And 2 o 'clock p .m.
05:29
Provided.
05:33
So this is also within, so let me claim this one...