A very crude model for the distribution of claim size, $X$, in a particular situation represents $X$ as a discrete random variable, which takes the values $£5,000, £10,000,$ and $£20,000$ with probabilities $0.4, 0.5,$ and $0.1$ respectively. Calculate the probability that of five randomly selected claims, three are for $£5,000$ each and the other two are for larger amounts.
Why will $X$ follow a binomial distribution? Don't I have to consider cases which contain $3$ of the $£5,000$ and different combinations of the other two?
Let $X_i$ be the claim amount of the $i$th selected claim $(i=1,2,3,4,5)$, and let's assume that the claim amounts within the five selections are independent. If the first three of your selection are for £$5000$ and the remaining two are not, then the event your looking at is $$E = \{X_1 = 5000, X_2 = 5000, X_3 = 5000, X_4 \neq 5000, X_5 \neq 5000\}$$ Using the independence of the $X_i$, you find $$P(E) = (.4)^3(1-.4)^2$$ Now, taking into consideration that we don't care which order we chose the three £$5000$ claims and the two non-£$5000$ claims, we have $\binom{5}{3}$ ways to select combinations of the claims. Therefore, we end up with the probability you're looking for as $$ \binom{5}{3} P(E) = \binom{5}{3} (.4)^3 (1-.4)^2. $$
Note that $X$ itself is not a binomial random variable. However, this problem boils down to considering a binomial random variable when you realize that each of the $X_i$ could have been interpreted as Bernoulli random variables with probability of success $.4$, since we only used the $X_i$ to distinguish between getting £$5000$ and not getting £$5000$. In this way, if you let $Y_i = 1$ if $X_i = 5000$ and $Y_i = 0$ if $X_i \neq 5000$, then you're looking for the probability $P(\sum_{i=1}^5 Y_i = 3)$, and the sum of independent identically distributed Bernoulli random variables is a Binomial random variable.