Uniform Distribution (probability) According to survey, a family spends between $\$800$ and $\$4300$ per year on all types of insurance in a particular city. Suppose the money spent is uniformly distributed between these amounts. If we randomly select $8$ families, what is the probability that at least two families spend more than $\$3600$ per year on insurance?
I have just learnt uniform distribution, but I am not sure how to find the probability if we randomly select from 8 families. Please help! 
1 - $\mathbb{P}(2 \text{ families spend more than } \$3600)$
 A: Hint to start the problem:
We begin by determining the probability that a single family spends more than $\$3600$ on insurance per year. Let us denote the amount a family spends on insurance by the random variable $X$, then we seek $\mathbb{P}(X\geq 3600)$, where $X$ is distributed uniformly on $(800,4300)$, i.e., $X \sim U(800,4300)$. Now, we know the cummulative distribution function for this uniform variable is $$F_X(x) = \frac{x-800}{4300-800} = \frac{x-800}{3500}.$$ Therefore, the $$\mathbb{P}(X\geq 3600) = 1 - \mathbb{P}(X \leq 3600) = 1 - F_X(3600)  = 1 - 0.8 = 0.2.$$ Now, let us denote the amount that the $8$ randomly selected families pay by the random variables $X_1,X_2,\ldots,X_8$. Because each of these random variables share the same distribution as $X$ from above, we have that for each $i \in \{1,2,\ldots,8\}$, that $\mathbb{P}(X_i \geq 3600) = 0.2$. Therefore, we can think of whether each family exceeds the threshold $\$3600$ like it is a Bernoulli trial (an unfair coin flip), with probability of success being $0.2$. Now, can you use the fact that these trials are independent (because the families are randomly selected), and identically distributed Bernoulli trials to answer your question? Have you learned about a Binomial distribution yet?
