Suppose a machine is recording a particle emission process of a piece of radioactive material.Once particles are emitted by the radioactive material, the machine will record immediately.It is known that the particles are emitted according to a Poisson process with an average of five per minute.
After recording a particle, the machine waits more than 30 seconds for the next particle emitted by the radioactive material. Conditioning on this, what is the probability that the radioactive material takes no more than one minute to emit the next particle?
What I know: $\mathbb{P}(wait <1 |w>30 s)=$?
Where conditional probability is $\mathbb{P}(B|A) = \mathbb{P}(B \cap A )/ \mathbb{P}(A)$
I know lambda must be divided by 2 since the time here is 30 and the rate is given for 5 minutes
Since it is measuring time in a poisson process I can use the CDF
$F(wait<1) = 1-e^{(-5)}$
$F(W>30) = 1- (1- e^{(-2.5)})$
But how do I find $\mathbb{P}(wait 30 s|wait < 1)$?
Am I interpreting this problem wrong?