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

Short Answer

Expert verified
The resulting probabilities are: \(P(C)=e^{-4}\), \(P(C^c)=1 - e^{-4}\), and \(P(C ∪ C^c)=1\)

Step by step solution

01

Compute P(C)

Compute the probability \(P(C)\) by integrating the given density function over the given interval [4, infinity). The definite integral is given by \(P(C) = \int_{4}^{\infty} e^{-x} dx\). It is noted that the integration of the function of the form \(e^{-x}\) from a to infinity is given by \(e^{-a}\). Therefore, \(P(C)=e^{-4}\)
02

Compute P(C^c)

Next, to compute the probability of the complement of C, \(P(C^c)\), note that \(C^c\) represents all possible events in the sample space that are not in set C. Thus, \(C^c = \{c: 0<c<4\}\). However, since events C and \(C^c\) are complementary and therefore mutually exclusive and exhaustive, i.e., they account for all possible outcomes, \(P(C) + P(C^c) = 1\). Therefore, we can solve for \(P(C^c) = 1 - P(C)\), hence \(P(C^c) = 1 - e^{-4}\)
03

Compute P(C ∪ C^c)

The union of C and \(C^c\) is the entire sample space. According to the rules of probability, the probability of the sample space is always 1, i.e., \(P(C ∪ C^c) = 1\)

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Key Concepts

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

Sample Space
In probability theory, the concept of a sample space is fundamental. The **sample space** represents all possible outcomes of a random experiment. For any given situation or scenario, it is crucial to define the sample space to understand all possible results that can occur.

For example, in our exercise, the sample space is \( \mathcal{C} = \{c: 0 < c < \infty\} \). This set includes all potential values of \(c\) that are greater than zero and continue indefinitely towards infinity. It's like considering any possible positive number, without any upper limit. This broad scope helps us understand the range of potential outcomes we might be dealing with.

When framing a probability problem, correctly identifying the sample space is a key initial step. It guides how we think about other sets derived from it, such as event spaces and complements.
Probability of Complement
The **probability of the complement** takes into account what happens when an event does not occur. If we know the probability of an event, calculating its complement is straightforward yet powerful.

Let's break it down: in our scenario, we are looking at set \(C = \{c: 4 < c < \infty\}\). The complement, \(C^c\), would thus be the **opposite** of this. So, \(C^c = \{c: 0 < c < 4\}\). This complement is everything in the sample space that is not in **C**.

An important rule in probability is that the sum of an event's probability and its complement equals 1. Therefore,
  • If you know \(P(C) = e^{-4}\), then \(P(C^c) = 1 - e^{-4}\).
Understanding complements allows us to quickly determine the probability of the rest of the sample space when we only know one part.
Probability Union Rule
The **Probability Union Rule** is a foundational principle in probability theory. It concerns the probability of either of two events occurring. This can often be seen in problems involving intersections or unions of sets.

In the context of our exercise, we consider the union \(C \cup C^c\). This union covers all elements in either set C or its complement, \(C^c\). It effectively spans the entire sample space. In our case, this means every element from \(0 < c < \infty\) is included.

A critical rule is that the probability of the entire sample space is always 1. So,
  • The probability \(P(C \cup C^c)\) will equal 1.
This concept is vital when you're piecing together probabilities for multiple events since it ensures all possibilities within the sample space are accounted for.

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Most popular questions from this chapter

Let \(X\) be a random variable of the continuous type that has pdf \(f(x)\). If \(m\) is the unique median of the distribution of \(X\) and \(b\) is a real constant, show that $$E(|X-b|)=E(|X-m|)+2 \int_{m}^{b}(b-x) f(x) d x$$ provided that the expectations exist. For what value of \(b\) is \(E(|X-b|)\) a minimum?

Consider the cdf \(F(x)=1-e^{-x}-x e^{-x}, 0 \leq x<\infty\), zero elsewhere. Find the pdf, the mode, and the median (by numerical methods) of this distribution.

Let \(X\) denote a random variable such that \(K(t)=E\left(t^{X}\right)\) exists for all real values of \(t\) in a certain open interval that includes the point \(t=1 .\) Show that \(K^{(m)}(1)\) is equal to the \(m\) th factorial moment \(E[X(X-1) \cdots(X-m+1)] .\)

A French nobleman, Chevalier de Méré, had asked a famous mathematician, Pascal, to explain why the following two probabilities were different (the difference had been noted from playing the game many times): (1) at least one six in 4 independent casts of a six-sided die; (2) at least a pair of sixes in 24 independent casts of a pair of dice. From proportions it seemed to de Méré that the probabilities should be the same. Compute the probabilities of \((1)\) and \((2)\).

(Monte Hall Problem). Suppose there are three curtains. Behind one curtain there is a nice prize while behind the other two there are worthless prizes. A contestant selects one curtain at random, and then Monte Hall opens one of the other two curtains to reveal a worthless prize. Hall then expresses the willingness to trade the curtain that the contestant has chosen for the other curtain that has not been opened. Should the contestant switch curtains or stick with the one that she has? If she sticks with the curtain she has then the probability of winning the prize is \(1 / 3 .\) Hence, to answer the question determine the probability that she wins the prize if she switches.

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