Probability that the millionth decay occurs within 100.2 seconds? 
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.
 A: $X$ 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.
A: 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 $.
A: 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
$$
\sum_{k=1000000}^\infty\frac{1002000^k}{k!}e^{-1002000}=0.9771959041
$$

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.
$$
%\frac1{\sqrt{2\pi}}\int_{-\frac{2000.5}{\sqrt{1002000}}}^{\infty}e^{-t^2/2}\,\mathrm{d}t=0.9771688952
$$
