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If \(P(E)=0.6\) and \(P(E \mid F)=0.34,\) are events \(E\) and \(F\) independent?

Short Answer

Expert verified
No, events E and F are not independent.

Step by step solution

01

- Understand the Independence Condition

For two events, E and F, to be independent, the occurrence of one should not influence the occurrence of the other. Mathematically, this means \[ P(E \text{ and } F) = P(E) \times P(F) \]. Alternatively, if \( P(E \mid F) = P(E) \), they are independent.
02

- Compare Given Probability

We are given that \( P(E) = 0.6 \) and \( P(E \mid F) = 0.34 \). According to the independence condition, if E and F were independent, it would be true that \( P(E \mid F) = P(E) \).
03

- Verify Independence

Compare the given probabilities. Since \( P(E \mid F) = 0.34 \) and \( P(E) = 0.6 \), and \( 0.34 \eq 0.6 \), events E and F are not independent.

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Key Concepts

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

independent events
In probability theory, determining whether events are independent is crucial. Two events, let's say E and F, are independent if the occurrence of one does not affect the occurrence of the other. This means the probability that event E happens is the same, regardless of whether event F occurs.

Mathematically, independence is expressed as:
\[ P(E \text{ and } F) = P(E) \times P(F) \].

Another way to check for independence is through conditional probability. If events E and F are independent, the conditional probability of E given F, denoted as \( P(E \text{ or } F) \), is equal to the probability of E. Therefore, if:
\[ P(E \text{ or } F) = P(E) \],
then the events are independent.

In our exercise, we are given \( P(E) = 0.6 \) and \( P(E \text{ or } F) = 0.34 \). Since \( P(E \text{ or } F) eq P(E) \), we can conclude that events E and F are not independent.
conditional probability
Conditional probability deals with the probability of an event occurring, given that another event has already happened. It is denoted as \( P(E \text{ or } F) \), which reads as the probability of event E occurring given that event F has occurred.

This can be calculated using the formula:
\[ P(E \text{ or } F) = \frac{P(E \text{ and } F)}{P(F)} \],
provided that \( P(F) eq 0 \). Understanding conditional probability is important in determining the relationship between events.

In our exercise, we are provided with \( P(E) = 0.6 \) and \( P(E \text{ or } F) = 0.34 \). By comparing \( P(E \text{ or } F) \) with \( P(E) \), we can analyze whether the occurrence of event F has an impact on the likelihood of event E. If they are equal, it indicates that event F does not influence event E, pointing towards independence.

But since 0.34 is not equal to 0.6, it demonstrates that F indeed affects E, thereby showing the dependence between these two events.
probability comparison
Comparing probabilities is a key tool in probability theory, especially for understanding the relationship between events. In the context of independence and conditional probability, comparing given probabilities helps determine whether events influence each other.

To ascertain if events E and F are independent, we compare \( P(E) \) and \( P(E \text{ or } F) \). If \( P(E \text{ or } F) = P(E) \), then E and F are independent. Conversely, if \( P(E \text{ or } F) eq P(E) \), the events are dependent.

In the exercise provided, we have:
  • \( P(E) = 0.6 \)
  • \( P(E \text{ or } F) = 0.34 \)
Clearly, \( 0.34 \) is not equal to \( 0.6 \). This result tells us that events E and F are not independent; event F affects the probability of event E happening.

This comparison is simple yet powerful in revealing the nature of the relationships between events, making it a fundamental step in solving many probability problems.

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Most popular questions from this chapter

For a parallel structure of identical components, the system can succeed if at least one of the components succeeds. Assume that components fail independently of each other and that each component has a 0.15 probability of failure. (a) Would it be unusual to observe one component fail? Two components? (b) What is the probability that a parallel structure with 2 identical components will succeed? (c) How many components would be needed in the structure so that the probability the system will succeed is greater than \(0.9999 ?\)

Fingerprints are now widely accepted as a form of identification. In fact, many computers today use fingerprint identification to link the owner to the computer. In \(1892,\) Sir Francis Galton explored the use of fingerprints to uniquely identify an individual. A fingerprint consists of ridgelines. Based on empirical evidence, Galton estimated the probability that a square consisting of six ridgelines that covered a fingerprint could be filled in accurately by an experienced fingerprint analyst as \(\frac{1}{2}\). (a) Assuming that a full fingerprint consists of 24 of these squares, what is the probability that all 24 squares could be filled in correctly, assuming that success or failure in filling in one square is independent of success or failure in filling in any other square within the region? (This value represents the probability that two individuals would share the same ridgeline features within the 24 -square region.) (b) Galton further estimated that the likelihood of determining the fingerprint type (e.g., arch, left loop, whorl, etc.) as \(\left(\frac{1}{2}\right)^{4}\) and the likelihood of the occurrence of the correct number of ridges entering and exiting each of the 24 regions as \(\left(\frac{1}{2}\right)^{8}\). Assuming that all three probabilities are independent, compute Galton's estimate of the probability that a particular fingerprint configuration would occur in nature (that is, the probability that a fingerprint match occurs by chance).

If the 2015 Hyundai Genesis has 2 engine types, 2 vehicle styles, 3 option packages, 8 exterior color choices, and 2 interior color choices, how many different Genesis's are possible?

In 1991 , columnist Marilyn Vos Savant posted her reply to a reader's question. The question posed was in reference to one of the games played on the gameshow Let's Make a Deal hosted by Monty Hall. Her reply generated a tremendous amount of backlash, with many highly educated individuals angrily responding that she was clearly mistaken in her reasoning. (a) Using subjective probability, estimate the probability of winning if you switch. (b) Load the Let's Make a Deal applet at www.pearsonhighered.com/sullivanstats. Simulate the probability that you will win if you switch by going through the simulation at least 100 times. How does your simulated result compare to your answer to part (a)? (c) Research the Monty Hall Problem as well as the reply by Marilyn Vos Savant. How does the probability she gives compare to the two estimates you obtained? (d) Write a report detailing why Marilyn was correct. One approach is to use a random variable on a wheel similar to the one shown. On the wheel, the innermost ring indicates the door where the car is located, the middle ring indicates the door you selected, and the outer ring indicates the door(s) that Monty could show you. In the outer ring, green indicates you lose if you switch while purple indicates you win if you switch.

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