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- Thread starter araz
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- Mar 5, 2012

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Hi araz, welcome to MHB!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.

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.

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Thank you Klaas. Did you use guess and check or numerical methods to get 0.824?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.

Regards

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- #4

- Mar 5, 2012

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I used an online calculator to find it.Thank you Klaas. Did you use guess and check or numerical methods to get 0.824?

Regards

Alternatively guessing can work, or we can look it up in a Poisson distribution table.