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Show that for any three events \(A, B\), and \(C\) with \(P(C)>0\), \(P(A \cup B \mid C)=P(A \mid C)+P(B \mid C)-P(A \cap B \mid C)\).

Short Answer

Expert verified
The identity is verified: \(P(A \cup B \mid C)=P(A \mid C)+P(B \mid C)-P(A \cap B \mid C)\) for any events with \(P(C) > 0\).

Step by step solution

01

Understand the Conditioned Probability Formula

Conditioned probability is defined as the probability of an event occurring given that another event has already occurred. For events \(A\) and \(C\), the conditioned probability \(P(A \mid C)\) is given by \(P(A \mid C) = \frac{P(A \cap C)}{P(C)}\). Similarly, for any event \(B,\) \(P(B \mid C) = \frac{P(B \cap C)}{P(C)}\) and \(P(A \cap B \mid C) = \frac{P(A \cap B \cap C)}{P(C)}\).
02

Express Union Event with Conditioning

To solve for \(P(A \cup B \mid C)\), we first express \(P(A \cup B)\) using the probability of the union formula: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). By conditioning on \(C\), we can write \(P(A \cup B \mid C) = \frac{P((A \cup B) \cap C)}{P(C)}\).
03

Use Distributive Property

Using the distributive property of intersection over union, \((A \cup B) \cap C = (A \cap C) \cup (B \cap C)\), the expression \(P((A \cup B) \cap C)\) becomes \(P((A \cap C) \cup (B \cap C))\).
04

Apply Union of Events Formula with Event C

Using the union of events formula again: \(P((A \cap C) \cup (B \cap C)) = P(A \cap C) + P(B \cap C) - P(A \cap B \cap C)\).
05

Substitute Back into Conditional Probability

Now, substitute this expression back into the conditional probability formula: \(P(A \cup B \mid C) = \frac{P(A \cap C) + P(B \cap C) - P(A \cap B \cap C)}{P(C)}\).
06

Simplify Using Conditional Probabilities

Divide each term in the numerator by \(P(C)\):\[ P(A \cup B \mid C) = \frac{P(A \cap C)}{P(C)} + \frac{P(B \cap C)}{P(C)} - \frac{P(A \cap B \cap C)}{P(C)}. \] Using the definition of conditional probability, this becomes:\[ P(A \cup B \mid C) = P(A \mid C) + P(B \mid C) - P(A \cap B \mid C). \]
07

Conclusion

We have shown that \(P(A \cup B \mid C)=P(A \mid C)+P(B \mid C)-P(A \cap B \mid C)\) holds true for any events \(A, B,\) and \(C\) with \(P(C)>0\).

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Key Concepts

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

Union of Events
The union of events in probability refers to the scenario where we are interested in the probability of either one event or another occurring, or both. This is represented by the symbol \(\cup\), meaning 'or'. For two events \(A\) and \(B\), the formula to find the probability of the union is given by:
  • \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
This formula accounts for the overlap where both events occur, which is represented by \(P(A \cap B)\). Subtracting this ensures that we don't count these events twice.
In our exercise, the union of events is modified by the presence of a conditioning event \(C\). The conditional union formula \(P(A \cup B \mid C)\) takes into account that we are interested in \(A \cup B\)'s occurrence given \(C\) has occurred, hence using the intersection with \(C\).
Understanding this concept helps in complex probability scenarios like predicting outcomes in games or making forecasts.
Probability Theory
Probability theory forms the mathematical foundation for quantifying uncertainty. It allows us to calculate the chance of various outcomes in different scenarios. In probability, the outcome space is called a sample space, and the likelihood of any outcome falling within a particular subset is called an event.
Key components of probability theory include events, which are the outcomes we are interested in, and their probability, often expressed as a number between 0 and 1.
  • A probability of 0 means an event will not occur.
  • A probability of 1 means it is certain to occur.
Conditional probability, a vital part of probability theory, measures the likelihood of an event given that another has occurred. It modifies the probability of an event by restricting the sample space, as seen in our solution where we condition on event \(C\). This adaptability makes probability theory fundamental in fields such as finance, weather forecasting, and decision-making processes.
Intersection of Events
The intersection of events in probability deals with finding the chance of two or more events happening simultaneously. It is denoted by \(\cap\) and is read as 'and'. For events \(A\) and \(B\), \(P(A \cap B)\) represents the probability of both events occurring at the same time.
  • For example, when tossing a fair dice, the event of rolling an even number and the event of rolling a number greater than three, their intersection would be \(\{4, 6\}\).
In the context of our exercise, the conditional intersection \(P(A \cap B \mid C)\) represents the probability that both \(A\) and \(B\) occur given \(C\) has occurred. Using intersections is crucial for dependencies between events, as they help us understand how the likelihood of one event affects another. Understanding these dependencies can clarify complex problems, such as assessing risks or evaluating compound probabilities in statistics and machine learning.

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Most popular questions from this chapter

Again consider a Little League team that has 15 players on its roster. a. How many ways are there to select 9 players for the starting lineup? b. How many ways are there to select 9 players for the starting lineup and a batting order for the 9 starters? c. Suppose 5 of the 15 players are left-handed. How many ways are there to select 3 left-handed outfielders and have all 6 other positions occupied by right-handed players? 34\. Shortly after being put into service, some buses manufactured by a certain company have developed cracks on the underside of the main frame. Suppose a particular city has 25 of these buses, and cracks have actually appeared in 8 of them. a. How many ways are there to select a sample of 5 buses from the 25 for a thorough inspection? b. In how many ways can a sample of 5 buses contain exactly 4 with visible cracks? c. If a sample of 5 buses is chosen at random, what is the probability that exactly 4 of the 5 will have visible cracks? d. If buses are selected as in part (c), what is the probability that at least 4 of those selected will have visible cracks?

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