A Geiger counter counts the number of alpha particles from radioactive

vadulgattp

vadulgattp

Answered question

2021-11-16

A Geiger counter counts the number of alpha particles from radioactive material. Over a long period of time, an average of 15 particles per minute occurs. Assume the arrival of particles at the counter follows a Poisson distribution.
a) Find the probability of exactly 20 particles arrive in a particular one minute period.
b) Find the probability of exactly one particle arrives in a particular one second period.
c) Find the probability that at least one particle arrives in a particular one second period.
d) Find the probability that at least two particles arrive in a particular 4 second period.

Answer & Explanation

Lounctirough

Lounctirough

Beginner2021-11-17Added 14 answers

According to the information presented, 15 particles are produced on average every minute.
Mean of poisson distribution (m)=15 
Process based on the probability mass function and the poisson distribution is,
P(X=x)=emmxx! 
a)The following equation can be used to calculate the likelihood that exactly 20 particles will arrive in one minute:
P(X=20)=e15152020! 
=0.04181 
Therefore, the likelihood that exactly 20 particles will arrive in one minute is 0.04181. 

b) The necessary probability that precisely one particle will arrive in one second can be calculated as:
One minute=15 particles 
60 seconds=15 particles 
1 second=0.25 particle 
So, here m=0.25 
P(X=1)=e0.250.2511! 
=0.19470 
As a result, the probability that exactly one particle will arrive in a given time interval is 0.19470. c) The necessary probability that at least one particle will arrive in a specific time interval can be calculated as:
P(X1)=1P(X=0) 
=1e0.250.2500! 
=0.80529 
Therefore, the likelihood that at least one particle will arrive in a specific one-second period is 0.80529. 

d) The necessary likelihood that at least two particles will arrive in a specific four-second period can be calculated as follows: in one second, 0.25 particles arrive, meaning that one particle will arrive in four seconds.
So, m=1 
P(X2)=1[P(X=0)+P(X=1)] 
=1[(e1100!)+(e1111!)] 
=0.26426 

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