# 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
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
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
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
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.