Question

If \(C_{1}, \ldots, C_{k}\) are \(k\) events in the sample space \(\mathcal{C}\), show that the probability that at least one of the events occurs is one minus the probability that none of them occur; i.e., $$P\left(C_{1} \cup \cdots \cup C_{k}\right)=1-P\left(C_{1}^{c} \cap \cdots \cap C_{k}^{c}\right)$$

Short answer

The stated equation \(P(C_{1} \cup \cdots \cup C_{k}) = 1 - P(C_{1}^{c} \cap \cdots \cap C_{k}^{c})\) is proven by using the complement rule, concepts of union and intersection of events, and the definition of not occurrence of all events equating to at least an occurrence in the sets.

Step-by-step solution

Understanding concepts

Understand the notation used - \(C_i\) represents any event from the sample space and \(C_i^c\) represents the complement of the event \(C_i\) which means that \(C_i\) does not occur. The union \(\cup\) of events implies that at least one of the events is occurring, while the intersection \(\cap\) implies that all the events are occurring simultaneously.

Complement Rule

Use the complement rule - the probability of an event not happening is 1 minus the probability of that event happening. In this context, \(1 - P(C_1^c \cap \cdots \cap C_k^c)\) represents the probability of the event that not all sets from \(C_1\) to \(C_k\) do not occur. If this event does not happen, it means that at least one of the events \(C_1\) to \(C_k\) is occurring, which is \(C_1 \cup \cdots \cup C_k\).

Final step

Finally, since the not occurrence of all events \(C_1\) to \(C_k\) equates to at least one occurrence in the sets from \(C_1\) to \(C_k\), it infers that \(P(C_1 \cup \cdots \cup C_k) = 1 - P(C_1^c \cap \cdots \cap C_k^c)\), hence proving the stated equation.