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Particles in a radioactive piece of material are decaying, and it is known that the number of decayed particles during a time period t (sec) is $Po(\lambda t)$-distributed. It is also known that the probability of there being at least one particle that has decayed during 60 seconds is 0.5. What value for $\lambda$ is the appropriate one to use in this model for radio active decay?

I am stumped on how to approach this problem. In particular I do not know how to make use of the given probability - P(at least one particle has decayed during a 60 second period)=0.5.

Eiraus
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  • Try to look at the complementary event "no particles decayed during a $60$-second period". What is its probability? How do you write it in terms of $\lambda$ and $t$? CAn you find $\lambda$ now? – TZakrevskiy Oct 19 '17 at 13:37

1 Answers1

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\begin{align} &\begin{cases} P(\text{at least one particle has decayed during a $60$ second}) = 0.5 \\ P(\text{at least one particle has decayed during a $60$ second}) = 1- P(\text{$0$ particle has decayed during a $60$ second}) \end{cases} \\ \Rightarrow &P(\text{$0$ particle has decayed during a $60$ second})= 1-0.5 = 0.5 \quad (1) \end{align}

By using $(1)$, we have \begin{align} &P(k \text{ events in 60 second}) = \frac{e^{-\lambda}\times \lambda^{k}}{k!} \Rightarrow \\ &P(0 \text{ events in 60 second}) = \frac{e^{-\lambda}\times \lambda^{0}}{0!} = 0.5 \Rightarrow e^{-\lambda} = 0.5 \Rightarrow \lambda = - ln(0.5) \end{align}

Hasan Heydari
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