91Ó°ÊÓ

A box contains the following four slips of paper, each having exactly the same dimensions: (1) win prize \(1 ;(2)\) win prize \(2 ;(3)\) win prize \(3 ;(4)\) win prizes 1,2 , and 3 . One slip will be randomly selected. Let \(A_{1}=\\{\) win prize 1\(\\}, A_{2}=\\{\) win prize 2\(\\}\), and \(A_{3}=\\{\) win prize 3\(\\} .\) Show that \(A_{1}\) and \(A_{2}\) are independent, that \(A_{1}\) and \(A_{3}\) are independent, and that \(A_{2}\) and \(A_{3}\) are also independent (this is pairwise independence). However, show that \(P\left(A_{1} \cap A_{2} \cap A_{3}\right) \neq P\left(A_{1}\right) \cdot P\left(A_{2}\right) \cdot P\left(A_{3}\right)\), so the three events are not mutually independent.

Step by step solution

01

Determine Probabilities for Individual Events

We know there are four slips, each equally likely, so each event has a probability of \(\frac{1}{4}\). Thus: - \(P(A_1) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\) (prizes 1 and 4 include prize 1),- \(P(A_2) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\) (prizes 2 and 4 include prize 2),- \(P(A_3) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\) (prizes 3 and 4 include prize 3).

02

Check Independence Between Pairs of Events

For pairwise independence, check if the probability of the intersection equals the product of their individual probabilities: - \(P(A_1 \cap A_2) = \frac{1}{4}\) (only prize 4), \(P(A_1) \cdot P(A_2) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\). Hence, \(A_1\) and \(A_2\) are independent.- \(P(A_1 \cap A_3) = \frac{1}{4}\) (only prize 4), \(P(A_1) \cdot P(A_3) = \frac{1}{4}\). Hence, \(A_1\) and \(A_3\) are independent.- \(P(A_2 \cap A_3) = \frac{1}{4}\) (only prize 4), \(P(A_2) \cdot P(A_3) = \frac{1}{4}\). Hence, \(A_2\) and \(A_3\) are independent.

03

Check Mutual Independence for All Three Events

For mutual independence, check if the probability of their intersection equals the product of their individual probabilities: - \(P(A_1 \cap A_2 \cap A_3) = \frac{1}{4}\) (only prize 4),- \(P(A_1) \cdot P(A_2) \cdot P(A_3) = \left(\frac{1}{2}\right)^3 = \frac{1}{8}\).Since \(\frac{1}{4} eq \frac{1}{8}\), the three events are not mutually independent.

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Key Concepts

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

Pairwise Independence

Pairwise independence in probability theory refers to a situation where any two events, out of a set of three or more events, are independent. This means that the occurrence of one event does not affect the probability of the other event happening. In mathematical terms, two events \(A\) and \(B\) are independent if \(P(A \cap B) = P(A) \cdot P(B)\).

In the context of our exercise with the slips, we examined combinations of winning prizes:

• For \(A_1\) and \(A_2\): the probability that both events happen (win prize 1 and prize 2) is \(\frac{1}{4}\), and since \(P(A_1) \cdot P(A_2) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\), it confirms independence.
• This same logic follows for \(A_1\) and \(A_3\), and \(A_2\) and \(A_3\).

This analysis demonstrates that each pair of these events are indeed pairwise independent. However, pairwise independence does not imply mutual independence, as we'll see in the following sections.

Mutual Independence

Mutual independence is a stricter condition than pairwise independence. Here, several events are mutually independent if every subset of these events is independent of the rest.

For three events \(A_1, A_2, A_3\), mutual independence requires:

• \(P(A_1 \cap A_2 \cap A_3) = P(A_1) \cdot P(A_2) \cdot P(A_3)\)
• Every pair of events and each event by itself should satisfy the conditions for pairwise independence

In the step-by-step solution, we calculated \(P(A_1 \cap A_2 \cap A_3) = \frac{1}{4}\), whereas \(P(A_1) \cdot P(A_2) \cdot P(A_3) = \left(\frac{1}{2}\right)^3 = \frac{1}{8}\).

Since \(\frac{1}{4} eq \frac{1}{8}\), the events are not mutually independent. This showcases that although pairs of events are independent, the trio fails the condition of mutual independence.

Conditional Probability

Conditional probability is the probability of an event occurring given that another event has already occurred. It’s denoted as \(P(A|B)\), representing the probability of event \(A\) under the condition that \(B\) has happened.

In our exercise, considering conditional probability could help understand why mutual independence fails despite pairwise independence.

• The odds of winning prize 1, for instance, might seem unaffected by knowing whether prize 2 or 3 is won.
• However, when considering all three events together, hidden dependencies often surface, hence defying mutual independence.

The key takeaway here is how conditional probability can serve as an insightful tool to dig deeper into the relationships between events.

Intersection of Events

The intersection of events in probability theory refers to situations where multiple events happen at the same time. The intersection of events \(A\) and \(B\), denoted \(A \cap B\), involves both events occurring simultaneously.

The calculation of the probability of intersections is crucial for understanding independence. In our exercise:

• The intersection \(A_1 \cap A_2\) stands for winning both prize 1 and prize 2 simultaneously, which had a probability of \(\frac{1}{4}\).
• Similarly, the intersections \(A_1 \cap A_3\) and \(A_2 \cap A_3\) also held the same probability, reflecting pairwise independence.
• However, \(A_1 \cap A_2 \cap A_3\) (winning all three prizes) resulting in \(\frac{1}{4}\) versus the calculated product of probabilities \(\frac{1}{8}\) highlights the lack of mutual independence.

Intersections demonstrate where dependencies or independencies exist among multiple events, making them a cornerstone concept in probability theory.