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For any events \(A\) and \(B\) with \(P(B)>0\), show that \(P(A \mid B)+P\left(A^{\prime} \mid B\right)=1 .\)

Short Answer

Expert verified
For events A and B with P(B)>0, P(A | B) + P(A' | B) = 1.

Step by step solution

01

Understand Conditional Probability

Conditional probability is defined as the probability of event \(A\) occurring given that event \(B\) has already occurred. It is represented as \(P(A \mid B)\) and is calculated using the formula \(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\).
02

Expressing the Complement

The complement of event \(A\), denoted as \(A'\), is the event that \(A\) does not occur. Therefore, the conditional probability of \(A'\) given \(B\) is \(P(A' \mid B) = \frac{P(A' \cap B)}{P(B)}\).
03

Use Probability Complements

Since \(A\) and \(A'\) are complements, \(A \cap B\) and \(A' \cap B\) are mutually exclusive events that together cover the entire sample space of event \(B\). Thus, \(P(A \cap B) + P(A' \cap B) = P(B)\).
04

Plug into Conditional Formula

Use the definition of conditional probability: \(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\) and \(P(A' \mid B) = \frac{P(A' \cap B)}{P(B)}\).
05

Sum of Conditional Probabilities

Add \(P(A \mid B)\) and \(P(A' \mid B)\):\[P(A \mid B) + P(A' \mid B) = \frac{P(A \cap B)}{P(B)} + \frac{P(A' \cap B)}{P(B)} = \frac{P(A \cap B) + P(A' \cap B)}{P(B)}\]Given from Step 3 that \(P(A \cap B) + P(A' \cap B) = P(B)\), we substitute:\[P(A \mid B) + P(A' \mid B) = \frac{P(B)}{P(B)} = 1\].
06

Conclusion

Thus, for any events \(A\) and \(B\) with \(P(B)>0\), it holds that \(P(A \mid B) + P(A' \mid B) = 1\).

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Key Concepts

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

Probability Complements
In probability theory, a **complement** refers to everything that is not part of a particular event. Consider an event \(A\), which occurs with probability \(P(A)\). The **complement of \(A\)**, denoted as \(A'\), is the scenario where \(A\) does not occur. A fundamental property of probability complements is that together, an event and its complement cover the entire sample space. Hence, the sum of their probabilities is always 1:
\[ P(A) + P(A') = 1 \].

When working with complements in conditional probability, it's important to realize that the same rule applies. If you're given a condition, like event \(B\) occurring, the probabilities of event \(A\) and its complement \(A'\) given \(B\) still add up to 1:
\[ P(A \mid B) + P(A' \mid B) = 1 \].

This formula is useful because it confirms that for any event, understanding its complement ensures you haven't missed anything in your analysis.
Complementary Events
**Complementary events** are pairs of events where one event is the complement of the other. If you consider a single event like flipping a coin, landing on heads and not landing on heads (which is tails) are complementary events. In other words, if one event happens, the other cannot.

For example, if you know that the probability of drawing a red card from a standard deck is \( \frac{1}{2} \), then the probability of drawing a non-red card (the complement) is also \( \frac{1}{2} \). Communicating this in terms of conditional probability:
  • If \(E\) is your event that occurs with \(P(E)\), then \(E'\) (the complement) occurs with \(P(E') = 1 - P(E)\).
  • For conditional probabilities, \(P(E \mid F)\) and \(P(E' \mid F)\) will sum to 1 under the condition \(F\).
  • This simple relationship between an event and its complement is powerful, allowing for quick insights into probability distributions.
Complementary events are also helpful in simplifying calculations, since knowing the probability of an event gives you the probability of its complement without additional computation.
Mutually Exclusive Events
**Mutually exclusive events** are events that cannot happen at the same time. In a roll of a single die, the event of rolling a three and the event of rolling a five are mutually exclusive. Only one of these outcomes can happen in a single roll. Understanding whether events are mutually exclusive is crucial in probability calculations.

In mathematical terms, if \(A\) and \(B\) are mutually exclusive events, then \(P(A \cap B) = 0\). This implies no overlap in the outcomes for these events. Consequently, the probability of either event \(A\) or \(B\) occurring is simply the sum of their probabilities: \(P(A \cup B) = P(A) + P(B)\).

When considering conditional probabilities and complements, mutual exclusivity plays a role too:
  • For complements \(A\) and \(A'\) of an event, \(A \cap A' = \emptyset\) meaning they can't both happen, e.g., mutually exclusive.
  • With conditional probabilities, this also means that if you're examining \(A\) given \(B\) and its complement, \(A'\) given \(B\), these states don't overlap either, and their sum confirms complete coverage.
Understanding mutually exclusive events helps in decomposing complex probability problems into simpler parts, ensuring all possible outcomes are considered.

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