Exercise of probability (Poisson process)

Question

Can you help me to solve this exercise? I do not understand how to solve some points :(. I have solved some items but I am in doubt about them..

It says:

A group of technicians provides technical support by telephone for CLEAR.Communication, addressing different issues. It was observed that these problems appear sporadically with an average rate of 3 problems every 7 hours, according to a Poisson process.

a)Calculate the probability that between 8 and 10 hours they will attend at least one problem, and in the following 6 hs they will attend 3 problems.

b)If between 9 hs and 17 hs they attended two problems. What is the probability that the first had occurred after 10 hs?

c)A technician goes to work at 9 o´clock and performs administrative tasks in the case of the first problem occurs after 11hs. Calculate the probability that in one week (5 working days) he has done at least 1 day administrative tasks.

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(NOT SURE)

a)λ = 3/7 (problems per hour)

Poisson process hypothesis:

P( k events during a time interval of length t ) = e^(-λt) * (λt)^k /k!,right?

Probability that between 8 o'clock and 10 o'clock they attend at least one problem, and in the following 6 hours they attend 3 problems:

= ( 1 - P( betw 8:00 and 10:00 there are 0 pbs) ) * e^(-λ6) * (6λ)^3 /3!

or

In 2 hours (between 8 and 10) , there are (3/7)(2) = 6/7 problems

L=6/7

P( at least one problem) = P( x >=1 ) = 1-P(x=0) = 1- e^(-6/7) =0.5756

In 6 hours , there are (3/7)(6) = 18/7 problems, L = 18/7

P(x=3) = e^(-18/7) (187)^3 / 3! = 0.2166

So, 0.5756 + 0.2166 = 0.7922

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(NOT SURE)

b)Events:

A = between 9 hrs and 17 hrs attended two problems

B = (between 9 and 17) the first occurred after 10 hrs

C = between 10 hrs and 17 hrs attended two problems

---->P( B | A) = P( B n A ) / P(A)

---->P(C) / P(A)

= e^(-7λ) * (7λ)^2 /2! * e^(-8λ) * (8λ)^2 /2!

or

There are 2 problems in 8 hours (between 9 and 17 hours)

In one hour, there are 8/2 = 4 problems, L = 4

P( after 10 hours) = P( x > 1) = 1-P(x=0) = 1- e^(-4) =0.9817

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c)I have no idea :S

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Can you help me?

Answer 1

1. Looks right.

2. The conditional distribution of the events on a given interval is uniform, thus the conditional distribution of the total number of events in $(10,17]$ given that in $[9,17]$ occurred $2$ events is $Bin(7/8, 2)$, hence, $$ P(T_1>10)=\left( \frac{17-10}{17-9} \right)^2 = \frac{7^2}{8^2}. $$

3. The probability that the first event occurs after $11$ is $$ P(T_1 > 2) = e^{-\lambda2}, $$ thus in $n=5$ days the distribution of the administrative tasks $(Y)$ is $Bin(e^{-\lambda2}, 5)$, hence $$ P(Y\ge 1 ) = 1 - P(Y=0) = 1 - (1- e^{-\lambda2})^5. $$

Answer 2

a) $ \lambda=3/7\text{ problems/hour}$ and $N(0)=0$

i) $ \Pr\{N(10)-N(8) \leqslant 1 \} ~{= 1 - \Pr\{N(10)-N(8) = 0 \} \\ = 1 - \Pr\{N(2) = 0 \} \\=1-\exp(-2\lambda)}$

ii) For homogeneous rate Poisson processes disjoint intervals are independent of each other. $ \Pr\{N(16){-}N(10^{+}){=}3\}~=~\exp(-6\lambda)(6\lambda)^{3}/3!$

b) ${\Pr\{N(10){-}N(9){=}0 \mid N(17){-}N(9){=}2\}}~{=\dfrac{\Pr\{N(10){-}N(9){=}0 , N(17){-}N(9){=}2 \}}{\Pr\{ N(17){-}N(9){=}2\}}\\=\dfrac{\Pr\{N(10){-}N(9){=}0\}\,\Pr\{N(17){-}N(10){=}2\} }{ \Pr\{ N(17){-}N(9){=}2\} } }$

Note that for a homogeneous Poisson process, if a number of arrivals are given in for a fixed interval order statistics of arrival pattern is uniform. Please see pg. 77-78 of http://www.rle.mit.edu/rgallager/documents/Poisson.pdf

c) The question is not posed in a clear manner regarding persistence of customers which are denied service (id est, is this process a homogeneous one or not?). Let us assume no persistence, this causes each day to become independent. Inverse of at least one day of administrative work is simpler, which is no administrative work in 5 days.

$\Pr(\text{No administrative work on day } i)~=~\Pr\{N_i(11){-}N_i(9){=}0\}$

$\Pr(\text{No administrative work on days between 1-5})~=~\Pr(\text{No administrative work on day } i)^{5}$

I hope this helps instead of you know giving you the fish. Please check the chapter in the link it is a nice resource.