Probability that the millionth decay occurs within 100.2 seconds?

Question

Radioactive decay of an element occurs according to a Poisson process with rate $10,000$ per second. What is the approximate probability that the millionth decay occurs within $100.2$ seconds?

Let $X$ be the number of decays and the number of expected decays within $100.2$ seconds is $\lambda=100.2\cdot10000=1002000.$ Thus $\bar{X}\sim \text{Poi}(1002000)$ and $\mu=1002000, \ \sigma=\sqrt{1002000}=1000.995.$

How to I formulate "probability that the millionth decay occurs within $100.2$ seconds?"

Is it $P(X>1000000)?$ I don't se how.

Answer 1

$X$ is equal to the number of decays over a given timespan of $100.2$ seconds. It could be any non-negative integer (theoretically).

If $X=3$ then $3$ decays occurred during than timespan. The fourth will then occur later. The same goes if $X<3$.

If $X\geq 4$, it means that there were a number of decays at least equal to $4$, hence the fourth was one of them.

It follows that $$P(X\geq 1000000)$$ is the correct formulation.

Answer 2

You have a Poisson process with expected value $1,002,000$. The chance that the millionth decay has not happened yet is the sum of the probabilities of exactly $0,1,2,\ldots ,999,999$ decays having happened. I believe you are supposed to use the normal approximation. Based on the figures you quote, a million decays is just about $2\sigma $ low so you need the chance that a random normal is greater than mean-$2\sigma $.

Answer 3

The expected number of decays in $100.2$ seconds is $1002000$. Thus, the Poisson distribution for exactly $k$ decays in $100.2$ seconds is $$ \frac{1002000^k}{k!}e^{-1002000} $$ The mean of this distribution is $1002000$, and the variance is $1002000$.

The probability of at least $1000000$ decays in $100.2$ seconds is $$ \sum_{k=1000000}^\infty\frac{1002000^k}{k!}e^{-1002000}=0.9771959041 $$

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The probability that a normally distributed random variable is greater than $\frac{2000.5}{\sqrt{1002000}}=1.998502496$ standard deviations less than the mean is approximately $0.9771688952$, which matches the exact value pretty closely. $$ %\frac1{\sqrt{2\pi}}\int_{-\frac{2000.5}{\sqrt{1002000}}}^{\infty}e^{-t^2/2}\,\mathrm{d}t=0.9771688952 $$