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An event \(F\) is said to carry negative information about an event \(E\), and we write \(F \searrow E\) if $$ P(E \mid F) \leq P(E) $$ Prove or give counterexamples to the following assertions: (a) If \(F \searrow E\), then \(E \searrow F\). (b) If \(F \searrow E\) and \(E \searrow G\), then \(F \searrow G\). (c) If \(F \searrow E\) and \(G \searrow E\), then \(F G \searrow E\). Repeat parts (a), (b), and (c) when \(\searrow\) is replaced by \(\lambda\), where we say that \(F\) carries positive information about \(E\), written \(F \nearrow E\), when \(P(E \mid F) \geq P(E)\)

Short Answer

Expert verified
In summary, we have shown that all three assertions (a), (b), and (c) are false by providing counterexamples to the inequalities in each case. For further practice, try similar analysis with the positive information concept, denoted by \(\nearrow\).

Step by step solution

01

Assuming the given condition

Let's assume that \(F \searrow E\) holds. That means \( P(E \mid F) \leq P(E) \)
02

Show if \(E \searrow F\) using the original definition

We must show whether the following is true or provide a counterexample: \( P(F \mid E) \leq P(F) \)
03

Write the conditional probability using the definition

We can write the given equation using the definition of conditional probability: \( P(F \mid E) = \frac{P(F \cap E)}{P(E)} \)
04

Analyze the inequality

Our goal is to check if the inequality holds: \( \frac{P(F \cap E)}{P(E)} \leq P(F) \) This inequality doesn't always hold, so a counterexample will suffice.
05

Counterexample

Let's assume the following probabilities: - \(P(E) = 0.4\) - \(P(F) = 0.6\) - \(P(E \cap F) = 0.29\) In this case, \(P(E \mid F) = 0.29/0.6 \approx 0.4833 \leq P(E)\), which means \(F \searrow E\). However, checking for the other inequality: \(P(F \mid E) = 0.29/0.4 \approx 0.725 \nleq P(F)\), which means \(E \nsearrow F\). So, the original assertion is false. (b) If \(F \searrow E\) and \(E \searrow G\), then \(F \searrow G\)
06

Assuming the given conditions

Let's assume that both \(F \searrow E\) and \(E \searrow G\) hold. That means: \( P(E \mid F) \leq P(E) \) and \( P(G \mid E) \leq P(G) \)
07

Show if \(F \searrow G\) using the original definition

We must show if the following is true (otherwise, provide a counterexample): \( P(G \mid F) \leq P(G) \)
08

Counterexample

Let's assume the following probabilities: - \(P(E) = 0.4\) - \(P(F) = 0.4\) - \(P(G) = 0.4\) - \(P(E \cap F) = 0.1\) - \(P(E \cap G) = 0.1\) - \(P(F \cap G) = 0.2\) In this case, \(P(E \mid F) = 0.1/0.4 = 0.25 \leq P(E)\), which means \(F \searrow E\), and \(P(G \mid E) = 0.1/0.4 = 0.25 \leq P(G)\), which means \(E \searrow G\). However, checking for the other inequality: \(P(G \mid F) = 0.2/0.4 = 0.5 \ngeq P(G) = 0.4\), which means \(F \nsearrow G\). So, the original assertion is false. (c) If \(F \searrow E\) and \(G \searrow E\), then \(F G \searrow E\)
09

Assuming the given conditions

Let's assume that both \(F \searrow E\) and \(G \searrow E\) hold. That means: \( P(E \mid F) \leq P(E) \) and \( P(E \mid G) \leq P(E) \)
10

Show if \(F G \searrow E\) using the original definition

We must show if the following is true (otherwise, provide a counterexample): \( P(E \mid F \cap G) \leq P(E) \) We can proceed directly to finding the counterexample.
11

Counterexample

Let's assume the following probabilities: - \(P(E) = 0.4\) - \(P(F) = 0.5\) - \(P(G) = 0.5\) - \(P(E \cap F) = 0.15\) - \(P(E \cap G) = 0.15\) - \(P(F \cap G) = 0.25\) - \(P(E \cap F \cap G) = 0.1\) In this case, \(P(E \mid F) = 0.15/0.5 = 0.3 \leq P(E)\), which means \(F \searrow E\), and \(P(E \mid G) = 0.15/0.5 = 0.3 \leq P(E)\), which means \(G \searrow E\). However, checking for the other inequality: \(P(E \mid F \cap G) = 0.1/0.25 = 0.4 \ngeq P(E) = 0.4\), which means \(F \cap G \nsearrow E\). So, the original assertion is false. We can similarly analyze the assertions when \(\searrow\) is replaced by \(\nearrow\). However, I'll leave that for you to practice! You can follow a similar approach using the definition of positive information and finding suitable examples.

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Key Concepts

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

Conditional Probability
Conditional probability is a fundamental concept that helps us determine the likelihood of an event occurring, given that another event has already occurred. It is denoted as \( P(A \mid B) \), which reads as "the probability of event \(A\) given event \(B\)." This is calculated using the formula:\[P(A \mid B) = \frac{P(A \cap B)}{P(B)}\]Here, \( P(A \cap B) \) represents the probability of both events \(A\) and \(B\) occurring. Conditional probability shines in understanding how information about one event can affect the likelihood of another.
Consider an example where you want to calculate the likelihood of it raining tomorrow (event \(A\)), given that it's cloudy today (event \(B\)). By observing the relationship between rain and cloudiness, you can use their joint probability to refine your estimations.
  • If \( P(A \mid B) = P(A) \), the events are independent; event \(B\) does not affect the probability of \(A\).
  • If \( P(A \mid B) > P(A) \), event \(B\) gives positive information about \(A\).
  • If \( P(A \mid B) < P(A) \), event \(B\) holds negative information regarding \(A\).
This impact is crucial when evaluating dependent events in real-world situations.
Event Relationships
Understanding event relationships can substantially affect how we interpret a probabilistic model. In probability theory, events can inform upon each other in positive or negative ways. To express these relationships, we use notations such as \( F earrow E \) and \( F \searrow E \).

Positive Information \( F earrow E \)

When \( F earrow E \), it denotes that the occurrence of \(F\) positively influences the probability of \(E\). This means that knowing \(F\) happened increases the likelihood of \(E\), described as \( P(E \mid F) \geq P(E) \).

Negative Information \( F \searrow E \)

Conversely, \( F \searrow E \) conveys that the occurrence of \(F\) has a negative effect on the probability of \(E\). Here, \( P(E \mid F) \leq P(E) \). Such negative relationships are counterintuitive and require careful analysis, often needing counterexamples to validate claims, as seen in the original exercise where assumptions didn't hold universally.
Deepening your understanding of these relationships helps in making smarter predictions and interpretations, especially in complex scenarios where events aren't independent.
Inequalities in Probability
Inequalities play a major role in probability theory, providing us with bounds and limitations on probabilities. When evaluating relationships between events, these inequalities offer insights about their interactions and dependencies.
One common inequality seen in conditional probabilities is establishing whether one event diminishes or enhances the likelihood of another. If you know \( P(E \mid F) \leq P(E) \), event \(F\) negatively impacts event \(E\). Similarly, it is crucial for real-world data analysis to test these inequalities through counterexamples. They can illustrate instances where these relationships don't follow assumptions, signifying the complexity of event interactions.
In solving probabilistic assertions, as with the exercise above, inequalities help to challenge or confirm the truth of the statements made about events. This critical process fosters a deeper understanding of how different events might behave under certain conditions, bridging theory with practice.
By applying these concepts and examining inequalities, you gain a holistic grasp of probability, useful for everything from academic studies to practical, data-driven decision-making.

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Most popular questions from this chapter

Urn A contains 2 white balls and 1 black ball, whereas urn \(B\) contains 1 white ball and 5 black balls. A ball is drawn at random from urn \(A\) and placed in urn \(B\). A ball is then drawn from um \(B\). It happens to be white. What is the probability that the ball transferred was white?

An urn initially contains 5 white and 7 black balls. Each time a ball is selected, its color is noted and it is replaced in the urn along with 2 other balls of the same color. Compute the probability that (a) the first 2 balls selected are black and the next 2 white; (b) of the first 4 balls selected, exactly 2 are black.

Three cards are randomly selected, without replacement, from an ordinary deck of 52 playing cards. Compute the conditional probability that the first card selected is a spade, given that the second and third cards are spades.

A friend randomly chooses two cards, without replacement, from an ordinary deck of 52 playing cards. In each of the following situations, determine the conditional probability that both cards are aces. (a) You ask your friend if one of the cards is the ace of spades and your friend answers in the affirmative. (b) You ask your friend if the first card selected is an ace and your friend answers in the affirmative. (c) You ask your friend if the second card selected is an ace and your friend answers in the affirmative. (d) You ask your friend if either of the cards selected is an ace and your friend answers in the affirmative.

A ball is in any one of \(n\) boxes. It is in the \(i\) th box with probability \(P_{i}\). If the ball is in box \(i\), a search of that box will uncover it with probability \(\alpha_{i}\). Show that the conditional probability that the ball is in box \(j\), given that a search of box \(i\) did not uncover it, is $$ \begin{array}{cl} \frac{P_{j}}{1-\alpha_{i} P_{i}} & \text { if } j \neq i \\ \frac{\left(1-\alpha_{i}\right) P_{i}}{1-\alpha_{i} P_{i}} & \text { if } j=i \end{array} $$

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