Find $\lim_{n\to \infty}\mathbb{P}(N_{I}(n)=k)$ which is a random color point in interval 
Divide the interval $[0,1]$ into $n$ equal-sized subintervals. Suppose that each endpoint of the intervals is colored red with probability $p_n=\lambda/n$ independent. For any interval $I\subset [0,1]$(possibly open/clsed...), let $N_{I}(n)$ be the number of red points in $I$. Find 
  $$\lim_{n\to \infty}\mathbb{P}(N_{I}(n)=k).$$
  How about the $$Cov(N_{I}, N_{J}), \forall I\cap J\neq 0$$

Based on the idea of @user125932:
Fix $I$ which has the number of endpoints $m$, then
$$\mathbb{P}(N_{I}(n)=k| I)=\binom{m}{k}(\frac{\lambda}{n})^k(1-\frac{\lambda}{n})^{m-k}$$
$$\mathbb{P}(N_{I}(n)=k)=\mathbb{E}(\mathbb{P}(N_{I}(n)=k| I))=\mathbb{E}\bigg(\binom{m}{k}(\frac{\lambda}{n})^k(1-\frac{\lambda}{n})^{m-k}\bigg)$$
As $n\to\infty, m=n|I|:=n_{I}$, we have
$$\lim_{n\to\infty}\mathbb{P}(N_{I}(n)=k)=\lim_{n\to\infty}\mathbb{E}\bigg(\binom{n_{I}}{k}(\frac{\lambda}{n})^k(1-\frac{\lambda}{n})^{n_{I}-k}\bigg)=?[\mathbb{E}(\frac{\lambda^k e^{-\lambda}}{k!})=\frac{\lambda^k e^{-\lambda}}{k!}]$$
But how about the next step?
 A: Let $m_n$ be the number of points in $I$ when $[0,1]$ is divided into $n$ equal intervals. It can be shown that $m_n=\lfloor bn\rfloor - \lceil an\rceil$. As you said, 
$$
P(N_I(n)=k)=\binom{m_n}k(\lambda/n)^k (1-\lambda/n)^{m_n-k}\tag1
$$ 
Now, we have the approximation $\binom{m_n}k\sim \frac{m_n^k}{k!}$, in the sense that 
$$
\lim_{n\to\infty}\frac{\binom{m_n}k}{\frac{m_n^k}{k!}}=1\tag2
$$
Can you prove this? In light of $(2)$, we can replace $\binom{m_n}{k}$ with $\frac{m_n^k}{k!}$ in $(1)$ without affecting the limit:
\begin{align}
P(N_I(n)=k)
&\sim \frac{m_n^k}{k!}(\lambda/n)^k(1-\lambda/n)^{m_n-k}\\
&=\left(\frac{m_n}n\right)^k\cdot \frac{\lambda^k}{k!}\cdot \left[\left(1-\frac{\lambda}n\right)^n\right]^{(m_n/n)}\cdot (1-\lambda/n)^{-k}
\end{align}
Now, we can simplify. We see the fraction $m_n/n$ appearing two places. It can be shown that $$\lim_{n\to\infty} \frac{m_n}n=b-a.$$
Furthermore, we have $\lim_n (1-\lambda/n)^n=e^{-\lambda}$ and $\lim_n (1-\lambda/n)^{-k}=1$. Making these replacements, we get
$$
\boxed{P(N_i(n)=k)\sim \frac{((b-a)\lambda)^k}{k!}\cdot e^{-(b-a)\lambda}.}
$$
A: The strong law of large numbers shouldn't help here, since that requires that your variables $X_i$ each have the same fixed distribution independent of $n$, but here, the distribution of each $X_i$ depends on $n$. 
A way forward: suppose there are $m$ endpoints in $I$. You can explicitly compute $\mathbb{P}(N_I(n) = k)$ in terms of $m, n, k, \lambda$ -- how many different ways can $k$ of the $m$ points be red? What is the probability of each such way? Then you can approximate $m$ in terms of $n$ (since intuitively, we should have $|I| \approx m/n$). This should allow you to find the limit as $n \to \infty$.
