A roadside assistance center receives 2 calls every half hour on average. a) Find the probability for this center to get at least 3 calls in two hours.
b) If the center received 3 calls in the first hour, find the probability for the center to receive 5 calls in the next hour.
This is clearly Poisson distribution. For a): 2 calls every half hour means 8 calls in two hours on average. Probability of at least 3 calls is $P(X \geq 3) = 1-P(X=0)-P(X=1)-P(X=2)$.
$\mathbb{P}\{X=0\} = e^{-\lambda} \frac{\lambda^0}{0!} = e^{-8} \frac{8^0}{0!}$ and similarly for the others, so I get $P=0.9862$.
How do I calculate the probability for the second case? I am a bit confused.
Edit: Since the two intervals are disjoint, the two probabilities will be independent, so we will calculate them and multiply. Is this so?
Thank you!