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Checking independence In each of three independent visits to a restaurant, you choose randomly between two of today's specials, TS1 and TS2, on tle memu. Let A denote \\{today's special on first visit is TS1\\}, B denote \\{ today's special on second visit is TS1\\}, C denote \\{ today's special on the first two visits are TS1\\}, and D denote \\{today's special on the three visits are TS1\\}. a. Find the probabilities of \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D}\). b. Which, if any, pairs of these events are independent? Explain.

Short Answer

Expert verified
A and B are independent events as are all other pairs like (A, C) or (B, D). P(A) = P(B) = 1/2, P(C) = 1/4, P(D) = 1/8.

Step by step solution

01

Understanding the Problem

To solve this exercise, we must determine the probabilities of choosing today's special, TS1, in three independent visits to a restaurant. Additionally, we must find which of these events are independent.
02

Calculate Probability of Event A

Since choosing between TS1 and TS2 is equally likely and independent for each visit, the probability of choosing TS1 on the first visit, event A, is \(P(A) = \frac{1}{2}\).
03

Calculate Probability of Event B

Similarly, the probability of choosing TS1 on the second visit, event B, is \(P(B) = \frac{1}{2}\).
04

Calculate Probability of Event C

Event C involves choosing TS1 on both the first and second visits, which is the intersection of events A and B. As the visits are independent, \(P(C) = P(A \cap B) = P(A) \cdot P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\).
05

Calculate Probability of Event D

Event D requires choosing TS1 on all three visits. This is an intersection of events A, B, and also choosing TS1 on the third visit. Therefore, \(P(D) = P(A) \cdot P(B) \cdot P(\text{Choose TS1 on third visit}) = \left(\frac{1}{2}\right)^3 = \frac{1}{8}\).
06

Determine Independence of Events

Two events are independent if \(P(A \cap B) = P(A) \cdot P(B)\). Since events A, B, and C are independent visits, \(P(A \cap B) = \frac{1}{4}\) is equal to \(P(A) \cdot P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\). Thus, events A and B are independent. We can verify similar calculations for other pairs, and since each visit is independent, all pairs (A, B), (A, C), (A, D), (B, C), (B, D), and (C, D) are independent.

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Key Concepts

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

Independence in Probability
In the world of probability, independence refers to the relationship between two events where the occurrence of one does not affect the probability of the other occurring. This concept is crucial when calculating the chance of multiple events happening together. If events A and B are independent, then the probability of both A and B happening is simply the product of their individual probabilities: \( P(A \cap B) = P(A) \cdot P(B) \).
For example, in our restaurant scenario, choosing today's special on any visit doesn’t depend on the choices made on previous or subsequent visits. Each choice is an independent event, meaning previous selections have no bearing on the outcome of the next selection. This independence simplifies our calculations because it allows us to compute compound probabilities simply by multiplying individual probabilities together.
Understanding independence is important in various fields, from business risk assessments to scientific experiments, as it helps in modeling situations where one action does not affect another.
Event Calculation
Event calculation involves determining the probability that specific outcomes occur. To calculate these probabilities, you must first identify each possible event and then determine the probability of these events.
In our example, each visit to the restaurant and the choice of TS1 or TS2 represents an individual event. To calculate the probabilities, we use the formula for independent events. Since the selection is equally likely for each visit, we have the following probabilities:
  • Probability of choosing TS1 on the first visit: \( P(A) = \frac{1}{2} \)
  • Probability of choosing TS1 on the second visit: \( P(B) = \frac{1}{2} \)
When calculating joint events like choosing TS1 on both the first and second visits, we use the intersection formula for independent events:
\( P(C) = P(A \cap B) = P(A) \cdot P(B) = \frac{1}{4} \).
Event D, being the selection of TS1 on all three visits, involves another multiplication step, illustrating how probabilities can cascade through several independent events:
\( P(D) = \left(\frac{1}{2} \right)^3 = \frac{1}{8} \).
Hence, understanding how to calculate these events accurately is crucial in probability, ensuring correct predictions and models.
Independent Events
Independent events are a foundational concept in probability theory. In practice, they represent scenarios where the outcome of one event provides no information about another. This is essential when analyzing situations with multiple possible outcomes, like our restaurant visits, because it simplifies the complexity of calculating probabilities.
In our example, the visits are separate and do not influence one another. Some pairs of events are particularly interesting:
  • The probability of choosing TS1 on both the first and second visits: Here, since the visits are independent, \( P(A \cap B) = P(A) \cdot P(B) \), confirming their independence.
  • Additionally, the independence extends to compound events. So, different combinations like (A, C) and (A, D) also hold the principle of independence since they adhere to the same multiplicative rule.
Understanding and identifying independent events allows for more straightforward calculations and aids in predicting outcomes without requiring excessive data or assumptions. It's a versatile concept that sees use from everyday decision-making to complex statistical modeling.

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