7. Promotion. The store "Crocus and Cactus" announced a promotion: if a customer buys three items, the cheapest one of them is free. The customer took ten items, all of different prices: 100 rubles, 200 rubles, 300 rubles, and so on - the most expensive item cost 1000 rubles.

a) (from 6th grade. 1 point). What is the smallest amount the customer will pay if they arrange their purchases skillfully before the cashier?

b) (from 6th grade. 1 point). What is the largest amount the customer will pay if they arrange the purchases poorly?

c) (from 9th grade. 3 points). Find the expected value of the amount the customer will pay if they arrange the purchases randomly.
Solution. a) The cheapest item in each trio should be as expensive as possible. If the buyer distributes the purchases into trios as follows (indicating only the prices):

$(1000,900,800),(700,600,500),(400,300,200)$ and a separate item for $100 \mathrm{p}.$, then he will pay $1000+900+700+600+400+300+100=4000$ rubles.

b) The worst distribution is when the buyer gets the three cheapest items for free under the promotion, which are 100, 200, and 300 rubles. In this case, the buyer will pay 4900 rubles. For example, the distribution could be:

$(1000,900,100),(800,700,200),(600,500,300)$ and a separate item for $400 \mathrm{p}$.

c) Let's take an item priced at $100 k$ rubles ( $k=1,2, \ldots, 10)$. Introduce a random variable - the indicator $I_{k}$, which is 1 if the buyer did not have to pay for this item, and 0 if they did. Obviously, $I_{10}=I_{9}=0$.

In other cases, $I_{k}=1$ only if, along with the item priced at $100 k$, two of the $10-k$ more expensive items are in the same trio. Such a combination can be formed in $C_{10-k}^{2}$ ways. The probability of such a combination is $\frac{C_{10-k}^{2}}{C_{9}^{2}}=\frac{(10-k)(9-k)}{72}$. The number $C_{9}^{2}$ in the denominator of the fraction is the total number of ways to select two items from the remaining nine to form a trio with the item priced at $100 k$.

We get the distribution

$$
I_{k} \sim\left(\begin{array}{cc}
0 \\
-\frac{k^{2}-19 k+18}{72} & \frac{k^{2}-19 k+90}{72}
\end{array}\right)
$$

The expected value: $\mathrm{E} I_{k}=\frac{k^{2}-19 k+90}{72}$. The total amount the buyer will save is

$$
S=100 I_{1}+200 I_{2}+\ldots+1000 I_{10}=100\left(I_{1}+2 I_{2}+\ldots+8 I_{8}\right)=100 \sum_{k=1}^{8} k I_{k}
$$

Transitioning to expectations, we get:

$$
\begin{gathered}
\mathrm{E} S=100 \sum_{k=1}^{8} \frac{k\left(k^{2}-19 k+90\right)}{72}= \\
=100\left(\frac{72}{72}+2 \cdot \frac{56}{72}+3 \cdot \frac{42}{72}+4 \cdot \frac{30}{72}+5 \cdot \frac{20}{72}+6 \cdot \frac{12}{72}+7 \cdot \frac{6}{72}+8 \cdot \frac{2}{72}\right)=916 \frac{2}{3} .
\end{gathered}
$$

The expected savings have been found. Therefore, the expected amount to be paid is

$$
\left.\frac{100+1000}{2} \cdot 10-916 \frac{2}{3}=5500-916 \frac{2}{3}=4583 \frac{1}{3} \text { (rubles }\right),
$$

which is 4583 rubles and 33 kopecks.

Answer: a) 4000 rubles; b) 4900 rubles; c) 4583 rubles 33.