11. In a box, there are 10 identical parts marked with numbers 1, 2, ..., 10. Six parts are randomly drawn. Find the probability that among the drawn parts there will be: a) part No. 1; b) parts No. 1 and No. 2.
Solution. a) The total number of possible elementary outcomes of the experiment is equal to the number of ways in which six parts can be drawn from ten, i.e., $C_{10}^{6}$.

Let's find the number of outcomes favorable to the event we are interested in: among the six selected parts, there is part 1, and consequently, the other five parts have different numbers. The number of such outcomes is obviously equal to the number of ways in which five parts can be selected from the remaining nine, i.e., $C_{9}^{5}$.

The desired probability is the ratio of the number of outcomes favorable to the event under consideration to the total number of possible elementary outcomes: $P=C_{9}^{5} / C_{10}^{6}=C_{9}^{4} / C_{10}^{4}=0.6$.

b) The number of outcomes favorable to the event we are interested in (among the selected parts, there are parts 1 and 2, so the other four parts have different numbers) is equal to the number of ways in which four parts can be drawn from the remaining eight, i.e., $C_{8}^{4}$. The desired probability $P=C_{8}^{4} / C_{10}^{6}=1 / 3$.