Question

1. In 2023, suppose one of the reactors of in the nuclear plant at Akkuyu starts to melt down. The emission of radioactive particles can be modeled as a Poisson process with (a) (5 points) What is the expected total number of particles that is emitted in the (b) (5 points) Suppose that exactly 300 particles emitted in the first two seconds. What rate A 100 particles/second. time window [30, 50] seconds is the distribution of the number of particles that emit in the next second? What is the expected value of the number of particles that emit in the next second? In order to stop the spread of radiation, the engineers but a set of n shields that can be either ON or OFF. Any OFF shield has no effect on the particle stream, whereas any ON shield blocks each particle with probability p, and letit through with probability 1-p, independently of all other particles. Suppose that each shield acts independently of all the other shields. (c) (10 points) For some fixed integer k (with 1 Sk ), suppose that exactly k of the shields are ON. What is the PDF and expected value of the number of particles that escape per second (d) (10 points) Suppose that the person in charge of turning ON the shield is a relative of a minister and he does not have the merits to do the job. He turns ON each shield with probability 0.5. Let U represent the number of particles that escape in the first second. Show that ElU 100(1-)"

Answer 1

The emission of radioactive particles can be modeled as a Poisson process is $\lambda =100 \textup{ particles/second}$

Poisson PMF is $P\left ( X=x\right )=e^{-\lambda }\frac{\left (\lambda \right ) ^x}{x!};x=0,1,2,3,....$ . The Poisson rate for $t=50-30=20$ seconds is $\lambda t$

a)The average number of particles in $t=20$ seconds is $\lambda 20$. Thus the expectation is

$\left ( 100 \right ) 20={\color{Blue} 2,000}$

b) Let $n$ be the number of particles emitted in the next 1 second given 300 particles emitted in the first 2 seconds.

Now the probability,

$P\left ( N\left ( 2+1 \right )=300+n|N\left ( 2 \right )=300 \right )=P\left ( N\left ( 3 \right )-N\left ( 2 \right )=300+n-300 \right )\\ P\left ( N\left ( 2+1 \right )=300+n|N\left ( 2 \right )=300 \right )=P\left ( N\left ( 3 \right )-N\left ( 2 \right )=n\right )\\$

Thus the PMF of $n$ is Poisson

${\color{Blue} P\left ( N=n\right )=e^{-\lambda }\frac{\left (\lambda \right ) ^n}{n!};n=0,1,2,3,....}$. The distribution of $N$ is Poisson with $\lambda =100 \textup{ particles/second}$

The average value is ${\color{Blue} \lambda =100}$

c) The probability that a shield passes a particle is $1-p$. The probability that $k$ number shields out of $n$ that passes particle is given by

$P\left ( X=k \right )=\binom{n}{k}\left ( 1-p \right )^kp^{n-k};k=0,1,2..,n$. The average number of particles that passes is given by

$100P\left ( X=k \right )={\color{Blue} 100\binom{n}{k}\left ( 1-p \right )^kp^{n-k};k=0,1,2..,n}$

d) The probability that a shield$1-0.5p$ passes a particle is  since the probability of switching on a shield is 0.5. The average number of shields that passes the particle is  $\left (1-p/2 \right )^n$. So number of The average number of particles that passes is given by

${\color{Blue} 100\left (1-p/2 \right )^n}$