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A sample space \(S\) consists of five simple events with these probabilities: $$\begin{array}{c}P\left(E_{1}\right)=P\left(E_{2}\right)=.15 \quad P\left(E_{3}\right)=.4\\\P\left(E_{4}\right)=2 P\left(E_{5}\right)\end{array}$$ a. Find the probabilities for simple events \(E_{4}\) and \(E_{5}\). b. Find the probabilities for these two events: $$A=\left\\{E_{1}, E_{3}, E_{4}\right\\}$$ $$B=\left\\{E_{2}, E_{3}\right\\}$$ c. List the simple events that are either in event \(A\) or event \(B\) or both. d. List the simple events that are in both event \(A\) and event \(B\).

Short Answer

Expert verified
Question: Find the probabilities for simple events \(E_4\) and \(E_5\), the probabilities for events \(A\) and \(B\), and list the simple events that are either in event \(A\) or event \(B\) or both, as well as the simple events that are in both event \(A\) and event \(B\). Answer: The probabilities for simple events \(E_4\) and \(E_5\) are 0.3 and 0.025, respectively. The probabilities for events \(A\) and \(B\) are 0.85 and 0.55, respectively. The simple events that are either in event \(A\) or event \(B\) or both are \(E_1\), \(E_2\), \(E_3\), and \(E_4\). The simple event that is in both event \(A\) and event \(B\) is \(E_3\).

Step by step solution

01

Determine the probabilities for simple events \(E_4\) and \(E_5\)

First, we will use the fact that the sum of probabilities of all simple events in a sample space equals 1: $$P(E_1)+P(E_2)+P(E_3)+P(E_4)+P(E_5)=1$$ Substitute the given probabilities: $$0.15+0.15+0.4+P(E_4)+2P(E_5)=1$$ Now, we have one equation with two unknowns. But we know that \(P(E_4)\) and \(P(E_5)\) are both greater than or equal to 0, so we can find their values by plugging in the lowest possible value for one of them and solving for the other one. Since both \(P(E_4)\) and \(P(E_5)\) are unknown, we need to start by finding ether \(P(E_4)\) or \(P(E_5)\).
02

Find the value of \(P(E_4)\)

For \(P(E_4)\), let's assume the lowest possible value for \(P(E_5)\) that is 0. Then: $$0.15+0.15+0.4+P(E_4)+2(0) =1$$ Solve for \(P(E_4)\): $$P(E_4)=1-0.15-0.15-0.4=0.3$$
03

Find the value of \(P(E_5)\)

Now that we have the value for \(P(E_4)\), we can plug it into our original equation to find \(P(E_5)\): $$0.15+0.15+0.4+0.3+2P(E_5)=1$$ Solve for \(P(E_5)\): $$2P(E_5)=1-0.15-0.15-0.4-0.3=0.05$$ Divide by 2: $$P(E_5)=0.025$$
04

Find the probabilities for events A and B

Event A consists of simple events \(E_1\), \(E_3\), and \(E_4\). To find the probability of event A, sum the probabilities of those simple events: $$P(A) = P(E_1) + P(E_3) + P(E_4) = 0.15 + 0.4 + 0.3 = 0.85$$ Event B consists of simple events \(E_2\) and \(E_3\). To find the probability of event B, sum the probabilities of those simple events: $$P(B) = P(E_2) + P(E_3) = 0.15 + 0.4 = 0.55$$
05

List the simple events in event A or event B or both

Simple events either in event A or event B or both are: $$E_1, E_2, E_3, E_4$$
06

List the simple events in both events A and B

The simple event that is in both event A and event B is: $$E_3$$

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Key Concepts

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

Sample Space
In probability, the concept of a sample space is like having a complete map of all possible outcomes for a given experiment. It's the foundation that allows us to calculate probabilities and make sense of what might happen when we conduct the experiment. The sample space is denoted by the symbol \( S \) and includes every possible event that can occur. For example, if you are rolling a six-sided die, the sample space is \( S = \{ 1, 2, 3, 4, 5, 6 \} \). Every possible number that the die can land on represents a "simple event." Understanding sample spaces is crucial because it ensures that we account for all possibilities, which is necessary when calculating the probabilities of specific events happening. Without knowing the sample space, it's like trying to solve a puzzle without all the pieces.
Simple Events
A simple event is a concept in probability that represents a single outcome within a sample space. Each simple event is distinct and doesn't depend on any other event's outcome. In our problem, the simple events are \( E_1, E_2, E_3, E_4, \) and \( E_5 \), each with its own probability indicating how likely it is to occur. Simple events are important building blocks in probability because they allow us to understand and predict the likelihood of more complex combinations of events. For instance, when we talk about event \( A \) consisting of \( E_1, E_3, \) and \( E_4 \), it is actually formed by combining these simple events. This helps in constructing probabilities for more compound events by systematically summing their components.
Event Probability Calculation
Calculating the probability of an event involves summing the probabilities of all simple events that comprise it. In probability theory, the likelihood of different events is expressed as a number between 0 and 1. Numbers closer to 1 indicate that an event is more likely to happen, whereas numbers near 0 suggest it is unlikely. For instance, to find the probability of event \( A \), which includes simple events \( E_1, E_3, \) and \( E_4 \), we add their respective probabilities: \[ P(A) = P(E_1) + P(E_3) + P(E_4) = 0.15 + 0.4 + 0.3 = 0.85 \] Similarly, the probability of event \( B \) consisting of \( E_2 \) and \( E_3 \) is calculated as: \[ P(B) = P(E_2) + P(E_3) = 0.15 + 0.4 = 0.55 \] These calculations allow us to determine the likelihood of more complex events occurring, making probability a highly predictive and useful tool.
Set Operations in Probability
Set operations are fundamental in probability as they help us define and calculate probabilities of combined events. In particular, union and intersection operations are crucial when dealing with probabilities. The **union** of two events, labeled as \( A \cup B \), represents all simple events that are in either \( A \) or \( B \) or both. In the exercise, the simple events from the union of \( A \) and \( B \) are \( E_1, E_2, E_3, \) and \( E_4 \). The **intersection** of two events, denoted by \( A \cap B \), contains only those simple events that are in both \( A \) and \( B \). Here, the intersection is simply \( E_3 \) as it is the only event common to both. These set operations simplify the calculation and understanding of probabilities for complex events, making them an indispensable part of probability theory.

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Most popular questions from this chapter

Accident records collected by an automobile insurance company give the following information: The probability that an insured driver has an automobile accident is .15 ; if an accident has occurred, the damage to the vehicle amounts to \(20 \%\) of its market value with probability \(.80,60 \%\) of its market value with probability \(.12,\) and a total loss with probability .08. What premium should the company charge on a \(\$ 22,000\) car so that the expected gain by the company is zero?

Evaluate these combinations: a. \(C_{3}^{5}\) b. \(C_{9}^{10}\) c. \(C_{6}^{6}\) d. \(C_{1}^{20}\)

The American Journal of Sports Medicine published a study of 810 women collegiate rugby players who have a history of knee injuries. For these athletes, the two common knee injuries investigated were medial cruciate ligament (MCL) sprains and anterior cruciate ligament (ACL) tears. \(^{8}\) For backfield players, it was found that \(39 \%\) had MCL sprains and \(61 \%\) had ACL tears. For forwards, it was found that \(33 \%\) had MCL sprains and \(67 \%\) had ACL tears. Since a rugby team consists of eight forwards and seven backs, you can assume that \(47 \%\) of the players with knee injuries are backs and \(53 \%\) are forwards. a. Find the unconditional probability that a rugby player selected at random from this group of players has experienced an MCL sprain. b. Given that you have selected a player who has an MCL sprain, what is the probability that the player is a forward? c. Given that you have selected a player who has an ACL tear, what is the probability that the player is a back?

A key ring contains four office keys that are identical in appearance, but only one will open your office door. Suppose you randomly select one key and try it. If it does not fit, you randomly select one of the three remaining keys. If it does not fit, you randomly select one of the last two. Each different sequence that could occur in selecting the keys represents one of a set of equiprobable simple events. a. List the simple events in \(S\) and assign probabilities to the simple events. b. Let \(x\) equal the number of keys that you try before you find the one that opens the door \((x=1,2,3,4)\). Then assign the appropriate value of \(x\) to each simple event. c. Calculate the values of \(p(x)\) and display them in a table. d. Construct a probability histogram for \(p(x)\).

Suppose the probability of remaining with a particular company 10 years or longer is \(1 / 6 .\) A man and a woman start work at the company on the same day. a. What is the probability that the man will work there less than 10 years? b. What is the probability that both the man and the woman will work there less than 10 years? (Assume they are unrelated and their lengths of service are independent of each other.) c. What is the probability that one or the other or both will work 10 years or longer?

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