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Chapter 1: Problem 3

Let the subsets \(C_{1}=\left\\{\frac{1}{4}

Short Answer

Expert verified
\(P_{X}(C_{1} \cup C_{2}) = \frac{5}{8}\), \(P_{X}(C_{1}^{c}) = \frac{7}{8}\), and \(P_{X}(C_{1}^{c} \cap C_{2}^{c}) = \frac{3}{8}\)

Step by step solution

01

Finding \( P_{X}(C_{1} \cup C_{2}) \)

Using the addition rule of probability, \( P_{X}(C_{1} \cup C_{2}) = P_{X}(C_{1}) + P_{X}(C_{2}) \). Given that \( P_{X}(C_{1}) = \frac{1}{8} \) and \( P_{X}(C_{2}) = \frac{1}{2} \), the sum of these two probabilities gives \( P_{X}(C_{1} \cup C_{2}) = \frac{5}{8} \).
02

Finding \( P_{X}(C_{1}^{c}) \)

The complement rule states that \( P(A^{c}) = 1 - P(A) \). Hence, \( P_{X}(C_{1}^{c}) = 1 - P_{X}(C_{1}) = 1 - \frac{1}{8} = \frac{7}{8} \).
03

Finding \( P_{X}(C_{1}^{c} \cap C_{2}^{c}) \)

We know that \( (C_{1}^{c} \cap C_{2}^{c}) = (C_{1} \cup C_{2})^{c} \). Hence, using the rule of complements, \( P_{X}(C_{1}^{c} \cap C_{2}^{c}) = P_{X}((C_{1} \cup C_{2})^{c}) = 1 - P_{X}(C_{1} \cup C_{2}) = 1 - \frac{5}{8} = \frac{3}{8} \).

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Key Concepts

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

Random Variables
In probability theory, a random variable is a fundamental concept referring to a variable whose possible values are numerical outcomes of a random phenomenon. Imagine tossing a dice; the number that appears is a random variable, because we cannot predict it with certainty. It could be any number ranging from 1 to 6.

Random variables can be continuous or discrete:
  • A discrete random variable has countable outcomes like the roll of a dice.
  • A continuous random variable, like in our exercise example, can take an infinite number of values within a given range.
The exercise given uses a subset of values between 0 and 1. Here, the random variable is continuous because it can take any value within this interval.
Addition Rule of Probability
The addition rule of probability allows us to find the probability that one of several events occurs. Specifically, if you have two or more events, you can calculate the probability of the union of these events occurring. This rule is especially useful when events are not mutually exclusive.

In the provided problem, we calculate the probability of events \( C_1 \) and \( C_2 \) occurring together. The formula is:
  • \( P(C_1 \cup C_2) = P(C_1) + P(C_2) \)
This formula works under the assumption that \( C_1 \) and \( C_2 \) are non-overlapping, meaning events do not intersect. If they did intersect, we should subtract the probability of their intersection. However, in the given exercise, \( C_1 \) and \( C_2 \) are distinct and the addition rule applies directly, giving us \( P(C_1 \cup C_2) = \frac{5}{8} \).
Complement Rule
The complement rule is a simple yet powerful tool in probability that helps find the probability of the "opposite" of a given event happening. Essentially, if you know the probability of an event \( A \), you can easily find the probability of the event not happening, known as \( A^c \) (A complement).

The formula is:
  • \( P(A^c) = 1 - P(A) \)
In our exercise, we used this rule to determine the likelihood that \( C_1 \) does not occur (\( C_1^c \)) and also that neither \( C_1 \) nor \( C_2 \) occur (\((C_1 \cup C_2)^c\)). For instance:
  • \( P(C_1^c) = 1 - \frac{1}{8} = \frac{7}{8} \)
  • \( P(C_1^c \cap C_2^c) = P((C_1 \cup C_2)^c) = 1 - \frac{5}{8} = \frac{3}{8} \)
This rule often simplifies solving probability problems, especially when dealing with the absence of results or events.

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Most popular questions from this chapter

The following game is played. The player randomly draws from the set of integers \(\\{1,2, \ldots, 20\\} .\) Let \(x\) denote the number drawn. Next the player draws at random from the set \(\\{x, \ldots, 25\\}\). If on this second draw, he draws a number greater than 21 he wins; otherwise, he loses. (a) Determine the sum that gives the probability that the player wins. (b) Write and run a line of \(\mathrm{R}\) code that computes the probability that the player wins. (c) Write an \(\mathrm{R}\) function that simulates the game and returns whether or not the player wins. (d) Do 10,000 simulations of your program in Part (c). Obtain the estimate and confidence interval, (1.4.7), for the probability that the player wins. Does your interval trap the true probability?

List all possible arrangements of the four letters \(m, a, r\), and \(y .\) Let \(C_{1}\) be the collection of the arrangements in which \(y\) is in the last position. Let \(C_{2}\) be the collection of the arrangements in which \(m\) is in the first position. Find the union and the intersection of \(C_{1}\) and \(C_{2}\).

Find the cdf \(F(x)\) associated with each of the following probability density functions. Sketch the graphs of \(f(x)\) and \(F(x)\). (a) \(f(x)=3(1-x)^{2}, 0

Let the sample space be \(\mathcal{C}=\\{c: 0

A random experiment consists of drawing a card from an ordinary deck of 52 playing cards. Let the probability set function \(P\) assign a probability of \(\frac{1}{52}\) to each of the 52 possible outcomes. Let \(C_{1}\) denote the collection of the 13 hearts and let \(C_{2}\) denote the collection of the 4 kings. Compute \(P\left(C_{1}\right), P\left(C_{2}\right), P\left(C_{1} \cap C_{2}\right)\), and \(P\left(C_{1} \cup C_{2}\right)\).
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