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Show that if \(A\) and \(B\) are independent, then so are \(A^{\prime}\) and \(B^{\prime}\) (assuming none of these events has zero probability). [Hint: \(A^{\prime} \cap B^{\prime}\) is the complement of \(A \cup B .\) ]

Short Answer

Expert verified
Given that events A and B are independent, we want to show that their complements, A' and B', are also independent. Using the hint \(A' \cap B'\) is the complement of \(A \cup B\), we can express the probability of the intersection of A' and B' as: \(P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)\) Since A and B are independent, we can express their union as: \(P(A \cup B) = P(A) + P(B) - P(A) * P(B)\) Substituting this expression into the complement probability, we have: \(P(A' \cap B') = 1 - P(A) - P(B) + P(A) * P(B)\) Recalling the complement rule, \(P(A') = 1 - P(A)\) and \(P(B') = 1 - P(B)\), we substitute these expressions: \(P(A' \cap B') = P(A') * P(B')\) Thus, if A and B are independent, then A' and B' are also independent.

Step by step solution

01

Recall the definition of independent events

Two events A and B are independent if and only if: \(P(A \cap B) = P(A) * P(B)\) We need to prove that: \(P(A' \cap B') = P(A') * P(B')\)
02

Express the probability of the intersection of complements using the hint

First, we need to express the probability of the intersection of A' and B' in terms of A and B using the hint given: \(P(A' \cap B') = P((A \cup B)')\) Now, we must express this probability using the formula for the complement probability: \(P((A \cup B)') = 1 - P(A \cup B)\)
03

Express the probability of the union of A and B using independence

Since A and B are independent, we can express the probability of their union using the formula: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) And, as A and B are independent, we can recall that: \(P(A \cap B) = P(A) * P(B)\) So, the probability of the union becomes: \(P(A \cup B) = P(A) + P(B) - P(A) * P(B)\)
04

Substitute the expression of the union probability into the complement probability

Now we can substitute the expression of \(P(A \cup B)\) into the complement probability: \(P(A' \cap B') = 1 - (P(A) + P(B) - P(A) * P(B))\)
05

Simplify the expression

Now, we can simplify the expression: \(P(A' \cap B') = 1 - P(A) - P(B) + P(A) * P(B)\)
06

Rewrite the probabilities of complements using the complement rule

Now, recall that: \(P(A') = 1 - P(A)\) \(P(B') = 1 - P(B)\) We can substitute these into the simplified expression: \(P(A' \cap B') = (1 - P(A)) * (1 - P(B))\) Which is equal to: \(P(A' \cap B') = P(A') * P(B')\)
07

Conclusion

Since the probability of the intersection of A' and B' is equal to the product of their individual probabilities, we have shown that: If A and B are independent, then A' and B' are also independent.

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Key Concepts

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

Understanding Probability Theory
Probability theory is a branch of mathematics concerned with the analysis of random phenomena. The central objects of probability theory are random events, stochastic processes, and variables, which we describe using concepts such as independence, complement, and intersection.

When we consider two events, like flipping a coin and rolling a die, probability theory helps us quantify how likely these events are to occur, both separately and in relation to each other. For instance, the likelihood of flipping a coin and getting heads (an event 'A') is unrelated to the roll of a die landing on six (an event 'B'), making these independent events. Their independence is mathematically expressed as: \(P(A \cap B) = P(A) \times P(B)\), where \(P\) denotes the probability and the intersection symbol \(\cap\) represents the occurrence of both events.

The significance of understanding these basic principles lies in their applicability to real-world situations where outcomes are uncertain, from simple games to complex scientific experiments and data analysis.
Complement of Events
In probability theory, the complement of an event is simply the occurrence of the event not happening. Represented by a prime symbol ('), it's mathematically defined for an event A as \(A'\)—the event that A does not occur.

Understanding the concept of the complement is pivotal to calculating probabilities because it broadens the analytical tools accessible to solve diverse problems. A fundamental relationship involving complements is given by the formula: \(P(A') = 1 - P(A)\). This equation is essential since finding the probability of an event not happening can be as straightforward as subtracting the probability of that event occurring from 1.

For example, if the probability of it raining tomorrow (event A) is 0.3, the probability of it not raining (event A') would be 1 - 0.3, which is 0.7. This complement rule simplifies many probability calculations and is directly used to prove the independence of complements as seen in the provided exercise.
Intersection of Events
The intersection of events refers to both events happening at the same time, denoted as \(A \cap B\). In the realm of probability, calculating the intersection is fundamental when determining how often two events occur together.

The concept of intersection interacts with the principle of independence; if two events are independent, the probability of their intersection is the product of their individual probabilities. This principle is used to answer more involved questions where two conditions must be satisfied simultaneously.

For example, if you want to find the likelihood of drawing a red card that's also a queen from a standard deck of cards, you'd multiply the probability of drawing a red card by the probability of drawing a queen, assuming the card's color doesn’t affect its rank.

If the exercises show the steps explicitly, it helps students recognize the structure behind probability problems and fosters a deeper comprehension. The exercise provided demonstrates these concepts by using the intersection in the context of complements to establish the independence of complement events.

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Most popular questions from this chapter

Auto Sales In April 2008, the probability that a randomly chosen new automobile was manufactured by General Motors was \(.21\), while the probability that it was manufactured by Toyota was. \(18 .{ }^{34}\) What is the probability that a randomly chosen new automobile was manufactured by neither company?

Based on the following table, which shows U.S. employment figures for 2007, broken down by educational attainment. \(^{49}\) All numbers are in millions, and represent civilians aged 25 years and over. Those classed as "not in labor force " were not employed nor actively seeking employment. Round all answers to two decimal places. Your friend claims that a person not in the labor force is more likely to have less than a high school diploma than an employed person. Respond to this claim by citing actual probabilities.

Concern the following chart, which shows the way in which a dog moves its facial muscles when torn between the drives of fight and flight. \({ }^{4}\) The "fight" drive increases from left to right; the "fight" drive increases from top to bottom. (Notice that an increase in the "fight" drive causes its upper lip to lift, while an increase in the "flight" drive draws its ears downward.) \(\nabla\) Let \(E\) be the event that the dog's flight drive is the strongest, let \(F\) be the event that the dog's flight drive is weakest, let \(G\) be the event that the dog's fight drive is the strongest, and let \(H\) be the event that the dog's fight drive is weakest. Describe the following events in terms of \(E, F, G\), and \(H\) using the symbols \(\cap, \cup\), and \(^{\prime} .\) a. The dog's flight drive is weakest and its fight drive is not weakest. b. The dog's flight drive is not strongest or its fight drive is weakest. c. Either the dog's flight drive or its fight drive fails to be strongest.

If \(B \subseteq A\) and \(P(B) \neq 0\), why is \(P(A \mid B)=1 ?\)

Auto Sales In April 2008, the probability that a randomly chosen new automobile was manufactured by Ford was \(.15\), while the probability that it was manufactured by Chrysler was .12. \({ }^{35}\) What is the probability that a randomly chosen new automobile was manufactured by neither company?

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