Although the correct answer is given, I'll give you some hints and a thought process here that may help you understand why the answer turns out the way it does.
The most important thing to arrive at the correct answer is to realize that you should be using the Binomial Distribution. How to see this?
The binomial distribution gives the probability of k successes in n independent yes/no experiments. In your case, the 'independent yes/no experiments' are the drawing of 10 families. That is, from a large population you randomly (i.e. the draws are independent) pick 10 families and look at their monthly expenses. One such draw is an experiment, and you make 10 of them.
Now, using the same naming conventions, we say that a family which expenses exceed \$3000 is a 'success' (remember, this is just a naming convention to be consistent with the language of the wikipedia article;)). The probability of 'success', denoted p in the article, is thus the probability that a uniform random variable on $[500,4500]$ is larger than 3000.
By reading about the binomial distribution you now know how to find the probability of finding exactly k families with expenses over $3000 in a random sample of 10 families. But the question asks about the probability of at least 2 such families out of 10. So, what do do? Note that the probability of at least 2 such families is the same as the probability of exactly 2 out of 10, plus the probability of exactly 3 out of 10, and so on all the way up to exactly 10 out of 10 such families.
However, you can also note that the complementary event of finding at least 2 such families is finding 0 or 1 such families. That is, if you do not find 2 or more such families in your 10 sampled, you must have found 0 or 1. So you can find the probability you seek by writing $$P(\text{At least 2 out of 10 families spend more than \$3000})=1 - P(\text{0 or 1 families spend more than \$3000})=1-P(\text{No family spends more than \$3000})-P(\text{1 family spends more than \$3000})$$Finally, you see that everything in this last row can be computed using the binomial distribution. I leave the actual computations to you.
Suppose the money spent is uniformly distributed between these amounts
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