Question

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

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

Write down the given information

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

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: \(0.15 + 0.15 + 0.4 + 2P(E_5) + P(E_5) = 1\)

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: \(P(E_5) = 0.1\)

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 \(P(E_4)\): \(P(E_4) = 2P(E_5) = 2(0.1) = 0.2\)

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

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

Key concepts

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.