Question

If \(C_{1}\) and \(C_{2}\) are subsets of the sample space \(\mathcal{C}\), show that $$ P\left(C_{1} \cap C_{2}\right) \leq P\left(C_{1}\right) \leq P\left(C_{1} \cup C_{2}\right) \leq P\left(C_{1}\right)+P\left(C_{2}\right) $$

Short answer

The relationship of the probability of the intersected subsets, individual subsets and the union of subsets can be demonstrated using the properties of set theory in a probability scenario. The inequalities \(P(C_{1} \cap C_{2}) \leq P(C_{1}) \leq P(C_{1} \cup C_{2}) \leq P(C_{1}) + P(C_{2})\) hold true, which validate the equations of the exercise.

Step-by-step solution

Apply Intersection Property of Probability

Firstly, analyze the left side of the inequality \(P(C_{1} \cap C_{2}) \leq P(C_{1})\). The intersection of two sets, \(C_{1} \cap C_{2}\), is a subset of each individual set. Therefore, the probability of the intersection is less than or equal to the probability of either individual set. This means that \(P(C_{1} \cap C_{2}) \leq P(C_{1})\).

Discussing Probability of Individual and Union Sets

Next, show \(P(C_{1}) \leq P(C_{1} \cup C_{2})\). By definition, the probability of an individual event is always less than or equal to the probability of that event or another. Thus, \(P(C_{1}) \leq P(C_{1} \cup C_{2})\).

Applying the Addition Rule of Probability

To show the right most part \(P(C_{1} \cup C_{2}) \leq P(C_{1}) + P(C_{2})\), you apply the addition rule of probability. This rule states that for any two events, the probability of the union of the events is less than or equal to the sum of the probabilities of the individual events. This rule holds even when the events are not mutually exclusive, hence \(P(C_{1} \cup C_{2}) \leq P(C_{1}) + P(C_{2})\).