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Assume that \(E\) and \(F\) are two events with positive probabilities. Show that if \(P(E \mid F)=P(E),\) then \(P(F \mid E)=P(F)\)

Short Answer

Expert verified
If \( P(E \mid F) = P(E) \), then \( P(F \mid E) = P(F) \).

Step by step solution

01

Understand Conditional Probability

To solve this problem, we need to understand what the statement means: \( P(E \mid F) = P(E) \) implies. This means the probability of event \( E \) occurring, given that \( F \) has occurred, is the same as the probability of \( E \) occurring without any conditions.
02

Use the Definition of Conditional Probability

Recall that the conditional probability \( P(E \mid F) \) is given by \( \frac{P(E \cap F)}{P(F)} = P(E) \). This is important to understand the relationship between \( E \) and \( F \).
03

Find \( P(E \cap F) \)

Using the equation from Step 2, express \( P(E \cap F) \) in terms of known probabilities: \[ P(E) \cdot P(F) = P(E \cap F) \] This equation tells us how likely both events \( E \) and \( F \) are to happen together.
04

Calculate \( P(F \mid E) \)

The conditional probability \( P(F \mid E) \) is \( \frac{P(E \cap F)}{P(E)} \). Substitute \( P(E \cap F) = P(E) \cdot P(F) \) into this formula: \[ P(F \mid E) = \frac{P(E) \cdot P(F)}{P(E)} \] Simplifying gives us \( P(F \mid E) = P(F) \).
05

Interpret the Conclusion

Since we derived that \( P(F \mid E) = P(F) \), this means the probability of \( F \) occurring, given \( E \) has occurred, is the same as the probability of \( F \) occurring without any condition. This confirms that if \( P(E \mid F) = P(E) \), then indeed \( P(F \mid E) = P(F) \).

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Key Concepts

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

Independence
In probability, when we say two events are independent, we mean that the occurrence of one event does not affect the occurrence of the other. For example, consider rolling a die and flipping a coin. These two events are independent because the result of the die roll does not influence the outcome of the coin flip, and vice versa.
In mathematical terms, independence between two events, say event \( E \) and event \( F \), is expressed using the equation:
  • \( P(E \, \text{and} \, F) = P(E) \cdot P(F) \)
This means that the joint probability of both events occurring is simply the product of their individual probabilities. If changes in the occurrence probability for one event have no impact on the probability of the other, they are independent.
The exercise we are considering shows us another important condition for independence: if \( P(E \mid F) = P(E) \), meaning the probability of event \( E \) given that \( F \) has occurred is the same as \( E \) occurring without any condition, then \( P(F \mid E) = P(F) \). This confirms the bidirectional nature of independence because the events do not influence one another.
Event Probability
Event probability is a fundamental concept in probability theory that deals with the likelihood of an event happening. An event, in simple terms, is any outcome or set of outcomes of a particular process. For instance, drawing a card from a deck or tossing a coin can each be considered an event.
The probability of an event is a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. For example:
  • The probability of drawing a red card from a standard deck of cards is \( \frac{26}{52} = 0.5 \), since half of the 52 cards are red.
  • The probability of rolling a 3 on a six-sided die is \( \frac{1}{6} \).
When working with probabilities, it's important to remember that the total probability of all possible outcomes should always sum to 1. This is often referred to as the sample space of possible events. In the context of the exercise, understanding event probability allows us to express conditional statements and relationships between events, making it crucial for deeper probability exploration.
Joint Probability
Joint probability refers to the probability of two events occurring simultaneously. It's a way of combining probabilities of individual events and is more complex than dealing with single events alone.
  • The joint probability of events \( E \) and \( F \) happening together is denoted as \( P(E \cap F) \).
This calculation is important because it gives insight into how two events interact with each other. In many practical situations, knowing the likelihood of events co-occurring can inform decisions and predictions.
In our exercise, the calculation of joint probability plays a central role. We see it in action when determining whether two events are independent. Recall from the provided solution steps that:
  • When \( P(E \mid F) = P(E) \) and subsequently \( P(E \cap F) = P(E) \cdot P(F) \), this setup confirms the independence of \( E \) and \( F \).
Joint probability thus forms a crucial part of evaluating conditional probabilities and the overall structure of complex probabilistic scenarios.

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Most popular questions from this chapter

A coin is tossed twice. Consider the following events. A: Heads on the first toss. \(B:\) Heads on the second toss. \(C:\) The two tosses come out the same. (a) Show that \(A, B, C\) are pairwise independent but not independent. (b) Show that \(C\) is independent of \(A\) and \(B\) but not of \(A \cap B\).

You are given two urns each containing two biased coins. The coins in urn I come up heads with probability \(p_{1}\), and the coins in urn II come up heads with probability \(p_{2} \neq p_{1}\). You are given a choice of (a) choosing an urn at random and tossing the two coins in this urn or (b) choosing one coin from each urn and tossing these two coins. You win a prize if both coins turn up heads. Show that you are better off selecting choice (a).

The Yankees are playing the Dodgers in a world series. The Yankees win each game with probability .6. What is the probability that the Yankees win the series? (The series is won by the first team to win four games.)

George Wolford has suggested the following variation on the Linda problem (see Exercise 1.2.25). The registrar is carrying John and Mary's registration cards and drops them in a puddle. When he pickes them up he cannot read the names but on the first card he picked up he can make out Mathematics 23 and Government \(35,\) and on the second card he can make out only Mathematics \(23 .\) He asks you if you can help him decide which card belongs to Mary. You know that Mary likes government but does not like mathematics. You know nothing about John and assume that he is just a typical Dartmouth student. From this you estimate: \(P(\) Mary takes Government 35\()=.5\) \(P(\) Mary takes Mathematics 23\()=.1\) \(P(\) John takes Government 35\()=.3\) \(P(\) John takes Mathematics 23\()=.2\) Assume that their choices for courses are independent events. Show that the card with Mathematics 23 and Government 35 showing is more likely to be Mary's than John's. The conjunction fallacy referred to in the Linda problem would be to assume that the event "Mary takes Mathematics 23 and Government \(35 "\) is more likely than the event "Mary takes Mathematics \(23 . "\) Why are we not making this fallacy here?

A card is drawn at random from a deck of cards. What is the probability that (a) it is a heart, given that it is red? (b) it is higher than a 10, given that it is a heart? (Interpret \(\mathrm{J}, \mathrm{Q}, \mathrm{K}, \mathrm{A}\) as \(11,12,13,14 .)\) (c) it is a jack, given that it is red?

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