91Ó°ÊÓ

To understand the problem, the student should first understand that the sample space \(\mathcal{C}\) is the set of all possible outcomes. \(C_{1}\) and \(C_{2}\) are subsets of the sample space, which means they are sets composed of possible outcomes. \(P(C)\) represents the probability of the event \(C\), and \(C_{1} \cap C_{2}\) and \(C_{1} \cup C_{2}\) refer to the intersection (common part) and the union (either or both) of \(C_{1}\) and \(C_{2}\) respectively.

Next, the student needs to prove the inequality \(P\left(C_{1} \cap C_{2}\right) \leq P\left(C_{1}\right) \leq P\left(C_{1} \cup C_{2}\right) \leq P\left(C_{1}\right)+P\left(C_{2}\right)\). To do this, the student can use the following reasoning: The probability of the intersection of two events is always less than or equal to the probability of either of the events - it cannot be greater because the intersection is part of each event. Therefore, \(P\left(C_{1} \cap C_{2}\right) \leq P\left(C_{1}\right)\). The probability of an event is always less than or equal to the probability of the union of that event with any other event - the union includes the event and possibly more. So, \(P\left(C_{1}\right) \leq P\left(C_{1} \cup C_{2}\right)\). Lastly, because the probability of the union of two events is always less than or equal to the sum of the probabilities of the two events (since the union includes both events but without double-counting the intersection), we can say that \(P\left(C_{1} \cup C_{2}\right) \leq P\left(C_{1}\right)+P\left(C_{2}\right)\).

This reasoning proves the given inequality to be true. The student should then summarize the steps and conclusions in the solution.