The discussion focuses on calculating the new emission rate of a radioactive source that emits particles following a Poisson distribution. Initially, the source emits at a rate of 1 particle per second. The probability of observing 0 or 1 emission in 4 seconds is given as 0.8, leading to the conclusion that the new average emission rate is approximately 0.206 particles per second. This calculation was confirmed using an online calculator, with alternative methods including guess and check or referencing a Poisson distribution table.
PREREQUISITESStudents in statistics, physicists working with radioactive materials, and professionals involved in risk assessment and safety analysis in nuclear industries will benefit from this discussion.
araz said: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 said: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 said:Thank you Klaas. Did you use guess and check or numerical methods to get 0.824?
Regards