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

Given the properties of probabilities, it is possible to show that \(P(C_{1} \cap C_{2}) \le P(C_{1}) \le P(C_{1} \cup C_{2}) \le P(C_{1}) + P(C_{2})\) by comparing each set and its respective subset or supersets.

Step-by-step solution

State Probability Properties

Recall two important properties of probability. The first is that the probability of an event is always non-negative, \(P(E) \ge 0\) for all subset \(E\) of the sample space \(\mathcal{C}\). The second important property is that the probability of the union of two events is the sum of their probabilities minus the probability of their intersection, which is denoted as \(P(E \cup F) = P(E) + P(F) - P(E \cap F)\).

Prove the First Inequality

We know \(C_{1} \cap C_{2}\) is a subset of \(C_{1}\). Therefore, \(P(C_{1} \cap C_{2}) \le P(C_{1})\) based on the property that the probability of an event is less than or equal to the probability of a larger event.

Show the Second Inequality

Similar to Step 2, we know \(C_{1}\) is a subset of \(C_{1} \cup C_{2}\). Therefore, \( P(C_{1}) \le P(C_{1} \cup C_{2})\) based on the same principle.

Prove the Third Inequality

We use the property of the union of two sets given in Step 1: \(P(C_{1} \cup C_{2}) = P(C_{1}) + P(C_{2}) - P(C_{1} \cap C_{2})\). Since \(P(C_{1} \cap C_{2}) \ge 0\), we have \(P(C_{1} \cup C_{2}) \le P(C_{1}) + P(C_{2})\).

Key concepts

Probability Properties

When diving into probability theory, we first need to understand some essential properties. One crucial property is that probabilities are always non-negative. This means for any event \(E\) from the sample space \(\mathcal{C}\), the probability \(P(E) \geq 0\). This ensures probabilities can never be negative, which aligns with the intuitive idea that an occurrence cannot have a negative chance.

Another fundamental concept is how probabilities behave when we look at unions of events. For two events \(E\) and \(F\), the probability of either event occurring, which is the union \(P(E \cup F)\), is calculated as the sum of the probabilities of each event minus the probability of both events occurring simultaneously: \(P(E) + P(F) - P(E \cap F)\). This formula helps us account for the overlap between events, ensuring we do not double-count the intersection.

Set Operations in Probability

In probability, events are often thought of as sets, and we use set operations to describe relationships between these events. One such operation is the intersection, denoted \(C_{1} \cap C_{2}\). This represents the event where both \(C_{1}\) and \(C_{2}\) occur. Given that intersections match the outcomes common to both events, it is naturally a subset of either event.

Another operation is the union, \(C_{1} \cup C_{2}\), which describes the occurrence of at least one of the events. The union contains all outcomes found in either \(C_{1}\) or \(C_{2}\). Because \(C_{1}\) is fully contained within \(C_{1} \cup C_{2}\), we have \(P(C_{1}) \leq P(C_{1} \cup C_{2})\). This concept supports the idea that including more outcomes (equivalently, more space) cannot reduce the probability.

• **Intersection** \((C_{1} \cap C_{2})\): both events occur.
• **Union** \((C_{1} \cup C_{2})\): either event occurs.

Probability Inequalities

Probability inequalities help us understand the bounds or limits of probabilities. One important inequality involving intersections is that for two events, \(P(C_{1} \cap C_{2}) \le P(C_{1})\). This agrees with the intuition that two events happening together should not be more probable than one event happening alone, as \(C_{1} \cap C_{2}\) is only part of \(C_{1}\).

We also have an inequality for unions: \(P(C_{1} \cup C_{2}) \le P(C_{1}) + P(C_{2})\). This ensures the probability does not exceed the sum of the probabilities of each event considered alone. Subtracting the intersection \(P(C_{1} \cap C_{2})\) in the union formula accounts for possible double-counting.

• **First Inequality**: \(P(C_{1} \cap C_{2}) \le P(C_{1})\)
• **Second Inequality**: \(P(C_{1} \cup C_{2}) \le P(C_{1}) + P(C_{2})\)