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 probability of event \(B=\left\\{E_{2}, E_{3}\right\\}\).

Short answer

Answer: The probability of event B is 0.55.

Step-by-step solution

Review the given information

The problem tells us the following probabilities: \(P(E_1) = P(E_2) = 0.15,\) \(P(E_3) = 0.4,\) \(P(E_4) = 2P(E_5).\)

Determine the additional probability

We know that the sum of all probabilities in a sample space must be 1: \(P(E_1) + P(E_2) + P(E_3) + P(E_4) + P(E_5) = 1.\) Substitute the given values into the equation: \(0.15 + 0.15 + 0.4 + 2P(E_5) + P(E_5) = 1.\) Combine the terms: \(0.7 + 3P(E_5) = 1.\)

Find the probability of \(E_5\)

Now, we can find the probability of \(E_5\): \(3P(E_5) = 1 - 0.7\) \(3P(E_5) = 0.3\) \(P(E_5) = \frac{0.3}{3}\) \(P(E_5) = 0.1\)

Calculate the probability of event \(B\)

Event B consists of \(E_2\) and \(E_3\). To find the probability of event B, we simply add the probabilities of \(E_2\) and \(E_3\): \(P(B) = P(E_2) + P(E_3)\) \(P(B) = 0.15 + 0.4\) \(P(B) = 0.55\) Thus, the probability of event \(B = \{E_2, E_3\}\) is 0.55.

Key concepts

Sample Space

In probability theory, a sample space is the set of all possible outcomes in a given experiment. Think of it as the foundation upon which probabilities are built. It includes every simple event that can possibly occur. For example, if you're flipping a coin, the sample space is \( \{\text{Heads, Tails}\} \). It lists all potential outcomes of a coin flip.

In our exercise, the sample space consists of five simple events: \(E_1, E_2, E_3, E_4,\) and \(E_5\). Each of these events represents a possible outcome. Together, they form the complete set of possibilities, ensuring that nothing is left out.

Understanding the sample space is crucial because it allows us to determine the likelihood of different events. It's the starting point for calculating probabilities, as each event within this space has a specific probability associated with it.

Simple Events

Simple events are individual outcomes within a sample space. They represent the most basic possible outcomes that cannot be broken down further. Each simple event in a sample space has its own probability which indicates how likely it is to occur.

In our specific example, the simple events are \(E_1, E_2, E_3, E_4,\) and \(E_5\). These can be thought of as the atomic level of all possible outcomes. For instance:

• \(P(E_1) = 0.15\)
• \(P(E_2) = 0.15\)
• \(P(E_3) = 0.4\)
• \(P(E_4) = 2P(E_5)\)
• \(P(E_5)\)

These events are each unique and independent, standing alone in terms of the probabilities associated with them. Simple events lay the groundwork for more complex calculations, such as determining the likelihood of combined events and total probabilities.

Event Probability

Event probability refers to the chance that a particular event occurs. It's calculated by summing the probabilities of all the simple events that make up the event. For example, if an event consists of two simple events, the probabilities of these two are added together to find the total probability of the composite event.

Our exercise task was to find the probability of event \(B = \{E_2, E_3\}\). Since \(E_2\) and \(E_3\) are part of the sample space, and their probabilities are given, we can calculate the probability of \(B\) by adding them:

• \(P(B) = P(E_2) + P(E_3)\)
• \(P(B) = 0.15 + 0.4\)
• \(P(B) = 0.55\)

This means there's a 55% chance that either \(E_2\) or \(E_3\) will occur, illustrating how event probability helps quantify the overall likelihood of complex outcomes.

Sum of Probabilities

A fundamental rule in probability theory is that the sum of all probabilities in a sample space must equal 1, or 100%. This makes sense because when considering all possible outcomes, one of them must happen. Hence, their total probability adds up to 1.

In our exercise, we were given probabilities for most of the simple events but had to deduce the rest. Using the equation:
\[ P(E_1) + P(E_2) + P(E_3) + P(E_4) + P(E_5) = 1 \] Allowed us to solve for the missing probability:\

• \(0.15 + 0.15 + 0.4 + 2P(E_5) + P(E_5) = 1\)
• Combine them to find \(3P(E_5) = 0.3\)
• Solve the equation \(P(E_5) = 0.1\)

This crucial concept ensures that our probability calculations are consistent and accurate, forming the basis for understanding how probabilities interact within a sample space.