Consider a n-worker equal lining framework where clients show up as per a Poisson

measure with rate A, where the assistance times are outstanding arbitrary factors with rate g, and

where any appearance discovering all workers occupied quickly withdraws without getting any assistance. On the off chance that an

appearance discovers all workers occupied, find

(a) the normal number of occupied workers found by the following appearance,

(b) the likelihood that the following appearance discovers all workers free,

(c) the likelihood that the following appearance finds precisely I ot the workers free.

Q26

Occasions happen as indicated by a Poisson cycle with rate A. Each time an occasion happens, we should

choose whether or not to stop, with our goal being to stop at the last occasion to happen preceding

some predetermined time T, where T > I/X.. That is, if an occasion happens at time t, O < T, and

we choose to stop, at that point we win ifthere are no extra occasions by time T, and we lose othenvise.

It we don't stop when an occasion happens and no extra occasions happen by time T, at that point we lose.

Likewise, it no occasions happen by time T, at that point we lose. Consider the procedure that stops at the main occasion

to happen after some fixed time s T

(a) utilizing this methodology, what is the likelihood of winning?

(b) What worth of s expands the likelihood of winning?

(c) Show that ones likelihood ofwinning when utilizing the first procedure with the worth of s

determined to some extent (b) is l/e.

Q27

A supply is known to nave around 5000 units of water in It.

It is being exhausted at a consistent pace of 1000 units each day.

The event ota precipitation is known to follow a Poisson cycle vvith a mean of 0.2 each day.

For a precipitation to add 5000 units to supply, the likelihood is 0.8.

For a precipitation to add 8000 units to supply, the likelihood is 0.2.

a)

The goal is to decide the likelihood of having a vacant supply following five days.

The likelihood work for a Poisson cycle is given by the recipe

Where, Ris the given mean unit time

t is the necessary time stretch.

k is the necessary number of occasions.

The supply at present is known to hold 5000 units ot water and is every day drained at a consistent

pace of 1000 units each day.

Accordingly, it will require 5 days to purge the repository going on like this.

Henceforth, for the supply to be vacant following 5 days, no precipitation ought to happen over it.

The mean pace of event of precipitation is 0.2 each day.

Decide the mean pace of precipitation for 5 days.

= 0.2*5

utilize this mean worth to decide the likelihood of event ot no precipitation in 5 days concurring

to Poisson measure.

0!

e-l(l)

1

5)

The goal is to discover the likelihood ot the repository being vacant at some point inside the following ten

days.

The supply at present is known to hold just shy of 5000 units of water and is day by day drained at a

steady pace of 1000 units each day.

A precipitation is known to add either 8000 units ot water or 5000 units of water to supply.

One approach to nave the repository void whenever insider 10 days is to not have any precipitation in first

5 days, which makes the supply void toward the finish of fifth day.

It the precipitation happens at any day inside the 10 days, the supply is filled again by either 5000

units or 8000 units.

It a precipitation happens and fills 8000 units of water on quickly, obviously supply won't ever be

void. Along these lines, this situation is completely precluded pinnacle our necessary case.

Ita precipitation happens and fills 5000 units of water on quickly, the absolute water in the supply will be

just shy of 10000 units. This water can be exhausted out in under 10 days.

Along these lines, regardless of whether one precipitation happens inside the initial 5 days, yet adds 5000 units ot water, the repository

will in any case be unfilled toward the finish of tenth day.

utilize the above data to make an articulation for likelihood of having a vacant supply at

any day inside 10 days.

void repository

whin I O days

void repository

wilhin 10 days

I Rain adding

no vain in

Or on the other hand 5000 units whenever

initial 5 days

in 5 days

no downpour in

I Rain in

Downpour adds

first 5days initial 5 days

5000units

The likelihood for a precipitation to add 5000 units of water is given to be 0.8.

void supply

inside I O days

Downpour adding 5000

Downpour adds

no downpour in

+ P units anyiimein x P

initial 5 days

5000 units

5 days

utilize the mean decided before, to decide the likelihood of event of no or one precipitation

in 5 days as per Poisson measure.

void supply

whin I O days

0.8

(1)0

x 0.8

0!

l!

xo.8

1

1

Q28

A viral straight DNA particle of length, say, 1 is frequently known to contain a certain "checked

position," witn the specific area ot this imprint being obscure. One way to deal with finding the

stamped position is to cut the atom by specialists that break it at focuses picked by a

Poisson measure witn rate A. It is then conceivable to decide the part that contains the

checked position. For example: allowing m to signify the area on the line ofthe stamped position:

at that point if signifies the last Poisson occasion time before m (oro if there are no Poisson occasions in [O,

m]), and RI signifies the principal Poisson occasion time after m (or 1 if there are no Poisson occasions in

[m, 11), at that point it WOUld be discovered that the checked position lies among Ll and RI. Find

(a) PILI = 0}

ay rehashing the former cycle on indistinguishable duplicates ot the DNA particle: we can zero

in on the area of the stamped position. In the event that the cutting strategy is used on n indistinguishable duplicates

ofthe atom, yielding the information Li, 8, I = 1, .

, n, at that point it follows that the stamped position lies

among L and R, where

L=maxLi, R

I

= min Ri

(e) Find E[R-LI, and in doing as such, show that E[RL]