Failures occur for a mechanical process according to a Poisson process. Failures are classified as either major or minor. Major failures occur at the rate of 1.5 failures per hour. Minor failures occur at the rate of 3.0 failures per hour.
(c) Find the probability that in 2 hours, at least two major failures occur or at least two minor failures occur.
My initial intuition is that I want to find the union of at least two major, at least two minor. The problem is that I don't know how to get the intersection.
For the probability of minor being greater than 1 by time 2 I have:
$P(N_2 > 1) = 1 - (6e^{-6} + e^{-6}) = .98265$
And probability of major:
$P(N_2 > 1) = 1 - (3e^{-3} + e^{-3}) = .80085$
I hope this doesn't seem too wrong or that I had the right idea. Thank you in advance.
Edit: I tried superpositioning the two independentent Poisson processes.
This gave $P(N_2 > 1) = \sum_{k=2}^{\infty} \frac{e^{-9}9^k}{k!} = 1- (9e^{-9} + e^{-9}) = .9988$$
This follows from the fact that $P(N_2 > 1) = 1- P(N_2 \le 1)$
This is super close to what the book has but is still off. My textbook claims the answer is .997
Edit 2: $N_2$ is the number of arrivals by time 2.
Let $N_1$ and $N_2$ be the number of minor accidents and major accidents we observe, respectively.
If the two events are considered to be independent then
$$P(N_1\gt 1 \cap N_2 \gt 1)=P(N_1\gt 1)\cdot P(N_2 \gt 1)$$
Then since for two events $A$ and $B$, $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$ we have
$$\begin{align*} P(N_1\gt 1 \cup N_2 \gt 1) &=P(N_1\gt1)+P(N_2\gt1)-P(N_1\gt 1)\cdot P(N_2 \gt 1)\\\\ &=0.98265+0.80085-(0.98265\cdot0.80085)\\\\ &\approx 0.9965 \end{align*}$$
which agrees with your book's answer.