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Chapter 3: Problem 6

Prove that if \(E_{1}, E_{2}, \ldots, E_{n}\) are independent events, then $$ P\left(E_{1} \cup E_{2} \cup \cdots \cup E_{n}\right)=1-\prod_{i=1}^{n}\left[1-P\left(E_{i}\right)\right] $$

Short Answer

Expert verified
To prove the given formula for the probability of the union of n independent events, we first define their complementary events and use the fact that they are also independent. We calculate the probability of all complementary events occurring simultaneously and apply De Morgan's law to rewrite this probability as the complement of the union of the original events. Finally, we replace the probabilities of complementary events with their expressions in terms of the original events' probabilities, obtaining the desired formula: $$ P\left(E_{1} \cup E_{2} \cup \cdots \cup E_{n}\right) = 1 - \prod_{i=1}^{n}\left[1-P\left(E_{i}\right)\right] $$

Step by step solution

01

Define the complementary events

For each event \(E_i\), define the complementary event \(E_i^c\) as the event that \(E_i\) does not occur. By definition, the probability of a complementary event is: $$ P\left(E_{i}^{c}\right) = 1 - P\left(E_{i}\right) $$
02

The probability of all complementary events occurring simultaneously

Now we want to find the probability that all complementary events \(E_1^c, E_2^c, \dots, E_n^c\) occur simultaneously. Since the events \(E_1, E_2, \dots, E_n\) are independent, their complementary events are also independent. Therefore, the probability of all complementary events occurring simultaneously is the product of their individual probabilities: $$ P\left(E_{1}^{c} \cap E_{2}^{c} \cap \cdots \cap E_{n}^{c}\right) = \prod_{i=1}^{n} P\left(E_{i}^{c}\right) $$
03

Apply De Morgan's Law to complementary events

Using De Morgan's law, we can rewrite the intersection of complementary events as the complement of the union of the original events: $$ P\left(E_{1}^{c} \cap E_{2}^{c} \cap \cdots \cap E_{n}^{c}\right) = P\left(\left(E_{1} \cup E_{2} \cup \cdots \cup E_{n}\right)^{c}\right) $$
04

Calculate the probability of the union of the events

Now, we want to find the probability of the union of the events \(E_1, E_2, \dots, E_n\). Since we calculated the probability of the complement of the union in Step 3, we can use the concept of complementary events to find the probability of the union: $$ P\left(E_{1} \cup E_{2} \cup \cdots \cup E_{n}\right) = 1 - P\left(\left(E_{1} \cup E_{2} \cup \cdots \cup E_{n}\right)^{c}\right) $$
05

Apply the obtained probabilities to the formula

We have found the relationship between the probability of the union and the probability of the intersection of complementary events in Step 3 and the result obtained in Step 2. Combine these results: $$ P\left(E_{1} \cup E_{2} \cup \cdots \cup E_{n}\right) = 1 - P\left(E_{1}^{c} \cap E_{2}^{c} \cap \cdots \cap E_{n}^{c}\right) = 1 - \prod_{i=1}^{n} P\left(E_{i}^{c}\right) $$
06

Replace the probabilities of complementary events with the given formula

Recall that \(P\left(E_{i}^{c}\right) = 1 - P\left(E_{i}\right)\). Replace this expression into the formula from Step 5: $$ P\left(E_{1} \cup E_{2} \cup \cdots \cup E_{n}\right) = 1 - \prod_{i=1}^{n}\left[1-P\left(E_{i}\right)\right] $$ This proves the given formula for the probability of the union of n independent events.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complementary Events
In probability theory, a foundational concept is that of complementary events. For any event E, we define its complementary event, often denoted as Ec, as the event that E does not occur. Visually, if we think of E as a part of a whole space, Ec is everything outside of E. It's crucial to understand that the probability of an event and its complement always add up to 1, as what happens will either be E or Ec. Mathematically, this is expressed as:

\[ P(E) + P(E^c) = 1 \]
Therefore, knowing the probability of an event occurring naturally informs us of the probability of it not occurring. This concept often simplifies calculations and is instrumental in proofs, such as showing the probability of the union of multiple events, as seen in our example exercise.
Independence of Events
When discussing multiple events in the context of probability, understanding the concept of independence is key. Two events A and B are said to be independent if the occurrence of one does not affect the probability of the occurrence of the other. The formal mathematical expression for the independence of two events is:

\[ P(A \text{ and } B) = P(A) \times P(B) \]
The concept expands to more than two events, as seen in the problem where multiple events E1, E2... are considered independent. The notion of independence allows us to multiply the probabilities of individual events to find the probability of their intersection. It is this powerful principle that underpins our approach to calculating the probability of the union of independent events by first considering their complements.
De Morgan's Laws
Within the realm of probability and set theory, De Morgan's laws are a pair of transformation rules that relate the union and intersection of sets through their complements. They state that the complement of the union of a collection of sets is the intersection of their complements, and vice versa. Formally, for events A and B, the laws are written as:

\[ (A \text{ or } B)^c = A^c \text{ and } B^c \]
and
\[ (A \text{ and } B)^c = A^c \text{ or } B^c \]
These laws are essential when working with probabilities of multiple events, as they allow us to rewrite probability expressions involving unions and intersections in a way that we can apply our knowledge of complement probabilities. This mathematical tool plays a crucial role in the exercise presented, as it helps us switch between the union and intersection of events and their complements to reach the desired probability expression.
Probability Theory
Probability theory is the mathematical framework that deals with the likelihood of events occurring within a given set of possibilities. It provides methods to quantify uncertainty and is extensively used in fields as diverse as statistics, finance, science, and philosophy. The very essence of probability is to determine how likely an event is to happen based on the outcomes of a random experiment.

Central concepts include random events, outcomes, sample spaces, and various measures such as conditional probability, independence, and theorems such as Bayes' theorem. Probability theory gives us a cohesive and formalized way to interpret and calculate the probabilities of complex events, building the foundation for statistical inference and decision-making under uncertainty. Our example involving the union of events isn't just an isolated mathematical problem鈥攊t's a part of the extensive toolbox that probability theory offers to deal with randomness in the world around us.

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Most popular questions from this chapter

You ask your neighbor to water a sickly plant while you are on vacation. Without water it will die with probability .8; with water it will die with probability .15. You are 90 percent certain that your neighbor will remember to water the plant. (a) What is the probability that the plant will be alive when you return? (b) If 'it is dead, what is the probability your neighbor forgot to water it?

For the \(k\)-out-of- \(n\) system described in Problem 63 , assume that each component independently works with probability \(\frac{1}{2}\). Find the conditional probability that component 1 is working given that the system works, when (a) \(k=1, n=2\); (b) \(k=2, n=3\).

There are 3 coins in a box. One is a two-headed coin; another is a fair coin; and the third is a biased coin that comes up heads 75 percent of the time. When one of the 3 coins is selected at random and flipped, it shows heads. What is the probability that it was the two-headed coin?

A total of 46 percent of the voters in a certain city classify themselves as Independents, whereas 30 percent classify themselves as Liberals and 24 percent as Conservatives. In a recent local election, 35 percent of the Independents, 62 percent of the Liberals, and 58 percent of the Conservatives voted. A voter is chosen at random. Given that this person voted in the local election, what is the probability that he or she is (a) an Independent; (b) a Liberal; (c) a Conservative? (d) What fraction of voters participated in the local election?

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