Poisson distribution Statistics emission millisecond interval

Question

It is known that a particular radiological emission has a Poisson distribution, with an average count of 3 per millisecond. The probability that during any two millisecond interval there will be 3 or more emission is:

My answer using the poisson distribution formula is the following

1 - (( e^-3 * 3^0 / 0!) + (e^-3 * 3^1 / 1!) + (e^-3 * 3^2 / 2!))

1 - 0.0497... + 0.149... + 0.22...

= 0.5768

However, I got it wrong for some reason, what is the correct answer and solution? I am thinking that the 2 millisecond interval plays a role in the question but not sure how.

I know the answer is either 0.938 or 0.062

Correct solution and answer?

Answer 1

It was asked about any two millisecond interval. We should multiply given average for one millisecond interval by $2$ and get Poisson distribution with average $6$.

Then $$P(X\geq 3)=1-P(X<3)=1-(P(X=0)+P(X=1)+P(X+2))= $$ $$=1-\left(e^{-6}\frac{6^0}{0!} + e^{-6}\frac{6^1}{1!} + e^{-6}\frac{6^2}{2!}\right)=0.938031196.$$