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The emission rate changes such that the probability of 0 or 1 emission in 4 seconds becomes 0.8. What is the new rate?

araz

New member
Dec 13, 2018
2
A radioactive source emits particles at an average rate of 1 pe second. Assume that the number of emissions follows a Poisson distribution. The emission rate changes such that the probability of 0 or 1 emission in 4 seconds becomes 0.8. What is the new rate? Thanks.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,713
A radioactive source emits particles at an average rate of 1 pe second. Assume that the number of emissions follows a Poisson distribution. The emission rate changes such that the probability of 0 or 1 emission in 4 seconds becomes 0.8. What is the new rate? Thanks.
Hi araz, welcome to MHB! ;)

We have:
$$P(\text{0 or 1 emission in 4 seconds})=0.8 \\ \implies
P(\text{0 emission in 4 seconds}) + P(\text{1 emission in 4 seconds}) = \frac{\lambda^0 e^{-\lambda}}{0!} + \frac{\lambda^1 e^{-\lambda}}{1!} = (1+\lambda)e^{-\lambda} = 0.8 \\ \implies
\lambda \approx 0.824
$$
So the average number of emissions in 4 seconds is $0.824$, which is $0.206$ per second.
 

araz

New member
Dec 13, 2018
2
Hi araz, welcome to MHB! ;)

We have:
$$P(\text{0 or 1 emission in 4 seconds})=0.8 \\ \implies
P(\text{0 emission in 4 seconds}) + P(\text{1 emission in 4 seconds}) = \frac{\lambda^0 e^{-\lambda}}{0!} + \frac{\lambda^1 e^{-\lambda}}{1!} = (1+\lambda)e^{-\lambda} = 0.8 \\ \implies
\lambda \approx 0.824
$$
So the average number of emissions in 4 seconds is $0.824$, which is $0.206$ per second.
Thank you Klaas. Did you use guess and check or numerical methods to get 0.824?
Regards
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,713
Thank you Klaas. Did you use guess and check or numerical methods to get 0.824?
Regards
I used an online calculator to find it.
Alternatively guessing can work, or we can look it up in a Poisson distribution table.