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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 \(E_{5}\).

Short Answer

Expert verified
Answer: The probabilities of simple events \(E_4\) and \(E_5\) are: - \(P(E_4) = 0.2\) - \(P(E_5) = 0.1\)

Step by step solution

01

Write down the given information

The given probabilities are: \(P(E_1) = P(E_2) = 0.15\) \(P(E_3) = 0.4\) \(P(E_4) = 2P(E_5)\)
02

Use the total probability rule

The total probability of all simple events in a sample space is 1, so we have: \(P(E_1) + P(E_2) + P(E_3) + P(E_4) + P(E_5) = 1\) Substitute the given probabilities into the equation: \(0.15 + 0.15 + 0.4 + P(E_4) + P(E_5) = 1\) Since \(P(E_4) = 2P(E_5)\), we can substitute this into the equation: \(0.15 + 0.15 + 0.4 + 2P(E_5) + P(E_5) = 1\)
03

Solve the equation for \(P(E_5)\)

Combine like terms and simplify the equation: \(0.7 + 3P(E_5) = 1\) Subtract 0.7 from both sides: \(3P(E_5) = 0.3\) Divide both sides by 3: \(P(E_5) = 0.1\)
04

Find \(P(E_4)\) using the given relationship

We found \(P(E_5) = 0.1\), and it's given that \(P(E_4) = 2P(E_5)\). So, we can find \(P(E_4)\): \(P(E_4) = 2P(E_5) = 2(0.1) = 0.2\)
05

Write down the probabilities for simple events \(E_4\) and \(E_5\)

The probabilities for simple events \(E_4\) and \(E_5\) are: \(P(E_4) = 0.2\) \(P(E_5) = 0.1\)

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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, a sample space is the set of all possible outcomes or results of an experiment. It is fundamental to setting up any probability problem. Imagine rolling a fair die; the sample space for this scenario consists of the numbers 1 through 6. Each outcome is possible, so they are part of the sample space.
In our original exercise, the sample space consists of five simple events, noted as \(E_1, E_2, E_3, E_4,\) and \(E_5\). Each event represents a possible outcome of an experiment. When defining a sample space:
  • Ensure it covers all possible individual outcomes.
  • Make sure events are mutually exclusive and collectively exhaustive.
The idea is to consider every conceivable result so that no outcome is omitted from the probability calculations. This forms the groundwork for the entire exercise, allowing us to analyze and determine the probabilities of the events involved.
Simple Events
Simple events are the most basic outcomes of a probability experiment. Each simple event represents one and only one outcome, like flipping a coin and getting heads. In the context of probability:
  • Simple events are atomic—they cannot be broken down into smaller parts.
  • They have a probability associated with them, which typically ranges between 0 and 1.
In our exercise, simple events \(E_1, E_2, E_3, E_4,\) and \(E_5\) have distinct probabilities assigned to them. For example, \(P(E_1) = P(E_2) = 0.15\) and \(P(E_3) = 0.4\). The probabilities associated with simple events can be derived from definitions or given relationships, like \(P(E_4) = 2P(E_5)\). Understanding simple events is crucial for calculating probabilities accurately, ensuring each event's probability is properly accounted for.
Total Probability Rule
The Total Probability Rule emphasizes that the sum of probabilities of all simple events in a sample space must be equal to 1. This rule is a fundamental principle in probability theory.
So why is this important? It provides a check to ensure that all possibilities have been accounted for in an experiment.
  • By summing all probabilities, you verify completeness and correctness of the probability distribution.
  • It serves as a balancing equation for solving unknown probabilities.
In the given problem, using the Total Probability Rule helps us define an equation: \[ P(E_1) + P(E_2) + P(E_3) + P(E_4) + P(E_5) = 1 \] Substituting known values into this equation allows the calculation of unknown probabilities. When \(P(E_4)\) and \(P(E_5)\) are initially unknown, the Total Probability Rule guides us to discover them, ensuring all probabilities sum to 1, thus maintaining the integrity of the probability distribution.

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Most popular questions from this chapter

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 group of research proposals was evaluated by a panel of experts to decide whether or not they were worthy of funding. When these same proposals were submitted to a second independent panel of experts, the decision to fund was reversed in \(30 \%\) of the cases. If the probability that a proposal is judged worthy of funding by the first panel is \(.2,\) what are the probabilities that: a. A worthy proposal is approved by both panels. b. A worthy proposal is disapproved by both panels. c. A worthy proposal is approved by one panel.

For the experiments, list the simple events in the sample space, assign probabilities to the simple events, and find the required probabilities. Three children are selected, and their gender recorded. Assume that males and females are equally likely. What is the probability that there are two boys and one girl in the group?

Evaluate the permutations. $$ P_{1}^{20} $$

A college student frequents one of two coffee houses on campus, choosing Starbucks \(70 \%\) of the time and Peet's \(30 \%\) of the time. Regardless of where she goes, she buys a cafe mocha on \(60 \%\) of her visits. a. The next time she goes into a coffee house on campus, what is the probability that she goes to Starbucks and orders a cafe mocha? b. Are the two events in part a independent? Explain. c. If she goes into a coffee house and orders a cafe mocha, what is the probability that she is at Peet's? d. What is the probability that she goes to Starbucks or orders a cafe mocha or both?

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