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Chapter 4: Problem 9

Experiment II A sample space contains seven simple events: \(E_{1}, E_{2}, \ldots, E_{7} .\) Suppose that \(E_{1}, E_{2}, \ldots, E_{6}\) all have the same probability, but \(E_{7}\) is twice as likely as the others. Find the probabilities of the events. $$ C=\left\\{E_{2}, E_{4}\right\\} $$

Short Answer

Expert verified
Question: In a sample space with seven simple events, E1, E2, ..., E7, the first six events have equal probabilities, and the probability for E7 is twice the probability of the other events. Determine the probabilities for each of these events. Answer: The probabilities for the seven events are: P(E1)=P(E2)=...=P(E6)=1/8, and P(E7)=1/4.

Step by step solution

01

Identify the equal probabilities for the first six events

Let \(p\) be the probability for each of the first six events, i.e., \(P(E_1)=P(E_2)=...=P(E_6)=p\). Since the probability of \(E_{7}\) is twice the probability of the other events, we have \(P(E_7) =2p\). Now, we know that the sum of all the probabilities in the sample space must be equal to 1. That is: $$ P(E_1)+P(E_2)+\cdots+P(E_6)+P(E_7)=1 $$
02

Set up the equation and solve for p

We can rewrite the equation using the probabilities we defined in step 1: $$ p + p + p + p + p + p + 2p=1 $$ This simplifies to: $$ 6p+2p=1 $$ Combining the like terms and solving for \(p\), we get: $$ 8p=1 $$ $$ p = \frac{1}{8} $$
03

Calculate the probabilities for each event

Using the value of \(p\) found in step 2, we can find the probabilities for each event: $$ P(E_1)=P(E_2)=\cdots=P(E_6)=\frac{1}{8} $$ $$ P(E_7)= 2p =2\cdot\frac{1}{8}=\frac{1}{4} $$
04

Verify that the sum of probabilities is equal to 1

Finally, we will check that the sum of all probabilities is equal to 1: $$ P(E_1) + P(E_2) + \cdots + P(E_6) + P(E_7) = 6\left(\frac{1}{8}\right) + \frac{1}{4} = \frac{3}{4} + \frac{1}{4} = 1 $$ The sum of probabilities is equal to 1, as expected. Therefore, the probabilities of the seven events are: $$ P(E_1)=P(E_2)=\cdots=P(E_6)=\frac{1}{8} $$ $$ P(E_7)=\frac{1}{4} $$

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Key Concepts

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

Sample Space
Understanding the sample space is crucial for grasping basic probability concepts. In probability theory, the sample space of an experiment is the set of all possible outcomes that can occur. Think of it as a collection of every conceivable scenario that the experiment could yield. For example, when rolling a dice, the sample space is \[ \{1, 2, 3, 4, 5, 6\} \.

In our exercise, the sample space is given by the seven simple events, \(E_{1}, E_{2}, \ldots, E_{7} \). These events cover all possible outcomes of the experiment in question. It's necessary that the sample space must be well-defined and complete, meaning it includes every outcome that could possibly occur in the experiment, and no outcome is left out.
Simple Events
Simple events are the basic building blocks of probability theory. A simple event is an outcome that cannot be broken down into simpler components within the context of an experiment. It's essentially a single outcome from the sample space. For example, drawing a single card from a deck of cards would be a simple event.

In our example, each \(E_{i}\) represents a simple event within the sample space. When dealing with simple events, it's possible to assign probabilities to them. The key is that the probabilities must be non-negative and sum up to one across the entire sample space. As per the exercise, except for \(E_{7}\), all other simple events have equal probability, labelled as \(p\), illustrating a foundational rule in probability where the total must always equal one.
Probability Distribution
A probability distribution assigns a probability to each possible outcome in the sample space in a way that the total probability across the sample space equals one. It describes how the total probability is distributed over the outcomes of a random experiment.

The step by step solution presented gives us a clear example of a discrete probability distribution, where probabilities are assigned to distinct simple events. By establishing \(p\) as the probability for six of the events and \(2p\) for the seventh, these probabilities collectively form the distribution. After solving for \(p\), we find that each of the first six events has a probability of \(\frac{1}{8}\) and the seventh event has a \(\frac{1}{4}\), exemplifying how the probability mass is allocated among the events.

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Most popular questions from this chapter

A single fair die is tossed. Assign probabilities to the simple events and calculate the probabilities. \(D:\) Observe a number less than 5

Suppose that a certain disease is present in \(10 \%\) of the population, and that there is a screening test designed to detect this disease if present. The test does not always work perfectly. Sometimes the test is negative when the disease is present, and sometimes it is positive when the disease is absent. The following table shows the proportion of times that the test produces various results. $$ \begin{array}{lcc} \hline & \text { Test Is Positive }(P) & \text { Test Is Negative }(N) \\ \hline \text { Disease } & .08 & .02 \\ \text { Present }(D) & & \\ \text { Disease } & .05 & .85 \\ \text { Absent }(D 9 & & \\ & & \\ \hline \end{array} $$ a. Find the following probabilities from the table: \(P(D), P\left(D^{c}\right), P\left(N \mid D^{c}\right), P(N \mid D)\) b. Use Bayes' Rule and the results of part a to find \(P(D \mid N)\) c. Use the definition of conditional probability to find \(P(D \mid N)\). (Your answer should be the same as the answer to part b.) d. Find the probability of a false positive, that the test is positive, given that the person is disease-free. e. Find the probability of a false negative, that the test is negative, given that the person has the disease. f. Are either of the probabilities in parts d or e large enough that you would be concerned about the reliability of this screening method? Explain.

A French restaurant offers a special summer menu in which, for a fixed dinner cost you can choose from one of two salads, one of two entrees, and one of two desserts. How many different dinners are available?

Experiment III A sample space consists of five simple events with \(P\left(E_{1}\right)=P\left(E_{2}\right)=.15, P\left(E_{3}\right)=.4,\) and \(P\left(E_{4}\right)=2 P\left(E_{5}\right) .\) Find the probabilities for simple events \(E_{4}\) and \(E_{5}\).

Refer to Exercise 33. Suppose you are interested in following two independent traits in snap peas-seed texture \((\mathrm{S}=\) smooth \(, \mathrm{s}=\) wrinkled \()\) and seed color \((\mathrm{Y}=\) yellow, \(\mathrm{y}=\) green \()-\) in a secondgeneration cross of heterozygous parents. Remember that the capital letter represents the dominant trait. Complete the table with the gene pairs for both traits. All possible pairings are equally likely. a. What proportion of the offspring from this cross will have smooth yellow peas? b. What proportion of the offspring will have smooth green peas? c. What proportion of the offspring will have wrinkled yellow peas? d. What proportion of the offspring will have wrinkled green peas? e. Given that an offspring has smooth yellow peas, what is the probability that this offspring carries one s allele? One s allele and one y allele?
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