Question

Suppose \(A\) and \(B\) are events in a sample space such that \(P(A)=1 / 4, P(B)=5 / 8\). and \(P(A \cup B)=3 / 4 .\) What is \(P(A \cap B) ?\)

Short answer

\(P(A \cap B) = \frac{1}{8}\).

Step-by-step solution

Understanding the problem

We are given probabilities of two events, \(A\) and \(B\), from a sample space. We need to find \(P(A \cap B)\) using the information provided: \(P(A)=\frac{1}{4}\), \(P(B)=\frac{5}{8}\), and \(P(A \cup B)=\frac{3}{4}\).

Applying the formula for union of events

We know the formula \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). We can rearrange this to find \(P(A \cap B)\) by substituting the given probabilities.

Substituting the values

Substitute the given values into the formula: \[ \frac{3}{4} = \frac{1}{4} + \frac{5}{8} - P(A \cap B) \]. To simplify, first convert \(\frac{1}{4}\) into eighths: \(\frac{1}{4} = \frac{2}{8}\).

Simplifying the equation

Now substitute the fractions into the equation: \[ \frac{3}{4} = \frac{2}{8} + \frac{5}{8} - P(A \cap B) \]. Convert \(\frac{3}{4}\) into eighths: \(\frac{3}{4} = \frac{6}{8}\). The equation becomes \(\frac{6}{8} = \frac{7}{8} - P(A \cap B)\).

Solving for \(P(A \cap B)\)

To find \(P(A \cap B)\), rearrange the equation: \[ P(A \cap B) = \frac{7}{8} - \frac{6}{8} \]. Simplify to get \(P(A \cap B) = \frac{1}{8}\).

Key concepts

Union of Events

In probability theory, the concept of the union of events is essential for understanding complex event combinations. The union of two events, say event \(A\) and event \(B\), denoted as \(A \cup B\), essentially means that we consider the outcomes where either event \(A\) occurs, or event \(B\) occurs, or both events occur.

This is a way of determining the breadth of possibility when multiple conditions can satisfy our requirements. For the formula:

• \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).

This formula includes the intersection of \(A\) and \(B\), which we subtract because we've added it twice; once in \(P(A)\) and once in \(P(B)\). It's crucial in situations where events might overlap. This ensures that double counting of shared outcomes is avoided. In the original exercise, we used this formula to find unknown probabilities by rearranging it to solve for \(P(A \cap B)\).

Intersection of Events

The intersection of events, denoted by \(A \cap B\), refers to the set of outcomes that are common to both events \(A\) and \(B\). It's similar to finding where two sets overlap in mathematics. In probability, this concept helps identify outcomes that satisfy all given conditions simultaneously. When you calculate \(P(A \cap B)\), you're determining the likelihood that both event \(A\) and event \(B\) happen at the same time. This concept is fundamental in solving problems where multiple criteria must be satisfied.

For example, in the original exercise, we were tasked with finding the intersection of two events given their individual probabilities and their union. Solving \(P(A \cap B) = \frac{1}{8}\) provides us with the probability that both conditions are met, meaning both events occur together, which is crucial for decision-making and analysis in real-world scenarios. It's invaluable in complex probability exercises that involve multi-layered events.

Sample Space

Sample space is a fundamental concept in probability theory. It refers to the set of all possible outcomes of a random experiment. If you think of flipping a coin, the sample space would be \(\{\text{Heads, Tails}\}\). It encompasses everything that can possibly happen when the experiment is conducted. In a more complex scenario, like rolling a die, the sample space is \(\{1, 2, 3, 4, 5, 6\}\). A sample space is represented symbolically by \(S\) and every possible event is a subset of this space. In the context of the original exercise, events \(A\) and \(B\) are subsets of a larger sample space that includes all possible outcomes. Understanding a sample space is crucial because it establishes the foundation upon which we calculate probabilities. It sets the stage for what events can happen and helps us calculate the probabilities and understand intersections and unions more effectively.

The sample space was implicitly understood in the solution since all probability calculations are based on it being complete and covering all eventualities. Learning to define a sample space is a vital skill in understanding any probabilistic analysis.