Calculate probability of poisson process conditionated to arrival time

Question

I'm solving a problem and at some point i have to compute the probabilitys $ P(S_2 < 3|N(5) = 6)$ where N(t) is a Poisson process with parameter 1.5, $S_n$ is the arrival time of the nth event.

My attempt:

$$ P(S_2 < 3|N(5) = 6)= 1- P(S_2 \ge 3|N(5) = 6)$$ $$= 1-P(N(3) < 2 |N(5) = 6)$$ $$= 1-\left[P(N(3) =0 |N(5) = 6) + P(N(3) =1 |N(5) = 6) \right] $$ Then using the conditional probability and independence it could be calculated but I´m not sure if what I've done is correct

Answer 1

You seek the probability that the second event occurs before time $3$ —i.e. at least 2 events occur within $(0..3]$,— when given that $6$ events have occurred within time interval $(0..5]$ .

So, indeed, you are correct:

$$\begin{align}\mathsf P(S_2{<}3\mid N(5){=}6)&=\mathsf P(N(3){\geq}2\mid N(5){=}6)\\[2ex]&=1-\mathsf P(N(3){<}2\mid N(5){=}6)\\[2ex]&=1-[\mathsf P(N(3){=}0\mid N(5){=}6)+\mathsf P(N(3){=}1\mid N(5){=}6)]\end{align}$$

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Then using the conditional probability and independence it could be calculated …

Yes.   You should readily recognise the (conditional) distribution for the count for events that occur within the interval $(0..3]$ among the $6$ events given to have occurred within $(0..5]$.

And of course $(N(3)\mid N(5){=}6)\sim\mathcal{Binomial}(6,3/5)$