Question

Generalize Exercise \(1.2 .5\) to obtain $$ \left(C_{1} \cup C_{2} \cup \cdots \cup C_{k}\right)^{c}=C_{1}^{c} \cap C_{2}^{c} \cap \cdots \cap C_{k}^{c} $$ Say that \(C_{1}, C_{2}, \ldots, C_{k}\) are independent events that have respective probabilities \(p_{1}, p_{2}, \ldots, p_{k} .\) Argue that the probability of at least one of \(C_{1}, C_{2}, \ldots, C_{k}\) is equal to $$ 1-\left(1-p_{1}\right)\left(1-p_{2}\right) \cdots\left(1-p_{k}\right) $$

Short answer

The expression \(\left(C_{1} \cup C_{2} \cup \cdots \cup C_{k}\right)^{c}=C_{1}^{c} \cap C_{2}^{c} \cap \cdots \cap C_{k}^{c}\) holds true by generalizing De Morgan's laws. The probability of at least one of the events \(C_1, C_2, ..., C_k\) happening is \(1 - \left(1-p_{1}\right)\left(1-p_{2}\right) \cdots\left(1-p_{k}\right)\).

Step-by-step solution

Define the Given

Let us consider \(C_1, C_2, ..., C_k\) are independent events with respective probabilities \(p_1, p_2, ..., p_k\). The goal is to show that the expression \(\left(C_{1} \cup C_{2} \cup \cdots \cup C_{k}\right)^{c}=C_{1}^{c} \cap C_{2}^{c} \cap \cdots \cap C_{k}^{c}\) holds true and to find the probability of at least one of these events occurring.

Apply De Morgan’s laws

De Morgan's laws from set theory states that the complement of the union of two sets is the intersection of their complements. If we generalize it for k events, it becomes: \(\left(C_{1} \cup C_{2} \cup \cdots \cup C_{k}\right)^{c}=C_{1}^{c} \cap C_{2}^{c} \cap \cdots \cap C_{k}^{c}\).

Calculate the Probability of none of the events

The event \(\left(C_{1} \cup C_{2} \cup \cdots \cup C_{k}\right)^{c}\) represents none of the events \(C_1, C_2, ..., C_k\) happening. This can also be written as all of the individual events not happening at the same time, which is \(C_{1}^{c} \cap C_{2}^{c} \cap \cdots \cap C_{k}^{c}\). Since the events are independent, the probability will be \(\left(1-p_{1}\right)\left(1-p_{2}\right) \cdots\left(1-p_{k}\right)\)

Find the probability of at least one event

The probability of at least one of these events happening is the complement of none of them happening. Therefore, it becomes \(1 - \left(1-p_{1}\right)\left(1-p_{2}\right) \cdots\left(1-p_{k}\right)\)