Problem 1 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Let \(A\) and \(B\) be two events in a sample space for which \(\mathrm{P}(A)=2 / 3\), \(\mathrm{P}(B)=1 / 6\), and \(\mathrm{P}(A \cap B)=1 / 9\). What is \(\mathrm{P}(A \cup B) ?\)

Step-by-Step Solution

Answer

\(\mathrm{P}(A \cup B) = \frac{13}{18}\).

Step 1: Understanding the Exercise

We are given events \(A\) and \(B\) with their respective probabilities \(\mathrm{P}(A) = \frac{2}{3}\), \(\mathrm{P}(B) = \frac{1}{6}\), and \(\mathrm{P}(A \cap B) = \frac{1}{9}\). Our task is to find \(\mathrm{P}(A \cup B)\), the probability that either event \(A\) or event \(B\) or both occur.

Step 2: Using the Formula for Union of Two Events

The formula to find \(\mathrm{P}(A \cup B)\) is given by:\[ \mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B) \]This formula accounts for the overlap counted twice when simply summing \(\mathrm{P}(A)\) and \(\mathrm{P}(B)\).

Step 3: Substitute Known Values

Substitute the known values of \(\mathrm{P}(A)\), \(\mathrm{P}(B)\), and \(\mathrm{P}(A \cap B)\) into the formula:\[ \mathrm{P}(A \cup B) = \frac{2}{3} + \frac{1}{6} - \frac{1}{9} \]

Step 4: Simplify the Expression

To simplify, find a common denominator for \(\frac{2}{3}\), \(\frac{1}{6}\), and \(\frac{1}{9}\), which is 18:\[ \mathrm{P}(A \cup B) = \frac{12}{18} + \frac{3}{18} - \frac{2}{18} \]Combine the fractions:\[ \mathrm{P}(A \cup B) = \frac{12+3-2}{18} = \frac{13}{18} \]

Step 5: Conclusion

The probability of either event \(A\), event \(B\), or both events occurring is \(\frac{13}{18}\).

Key Concepts

Union of Events

In probability theory, the union of two events, denoted as \(A \cup B\), refers to the occurrence of either Event A, Event B, or both. It essentially covers all outcomes that belong to at least one of the events. To calculate the probability of the union of events \(A\) and \(B\), we use the formula:\[\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B)\]Let's break down the formula into manageable parts:

• The term \(\mathrm{P}(A) + \mathrm{P}(B)\) adds both probabilities as if there were no overlap.
• Since the overlap (\(\mathrm{P}(A \cap B)\)) is counted twice in the first sum, we need to subtract it once.

This formula neatly corrects for any double counting and provides the exact probability of either or both events occurring.
For example, if \(\mathrm{P}(A) = \frac{2}{3}\), \(\mathrm{P}(B) = \frac{1}{6}\), and \(\mathrm{P}(A \cap B) = \frac{1}{9}\), the probability \(\mathrm{P}(A \cup B)\) can be calculated by substituting these values into the equation and solving as demonstrated in the solution.

Sample Space

The sample space in probability theory encompasses all possible outcomes of a given experiment. It's the foundation for defining events, which are simply subsets of the sample space. To visualize, consider a simple dice roll:

• The sample space is represented by \(\{1, 2, 3, 4, 5, 6\}\), capturing every possible result of the roll.

Every event related to this dice roll is a subset of this space, whether it's rolling an even number \(\{2, 4, 6\}\) or a specific number like \(\{5\}\). In more complex examples, like card games or random draws, the sample space might involve numerous combinations or permutations.
For the problem involving events \(A\) and \(B\), their probabilities originate from this overall sample space, with \(\mathrm{P}(A)\), \(\mathrm{P}(B)\), and \(\mathrm{P}(A \cap B)\) directly relating to the occurrences within it.

Intersection of Events

When discussing probability, the intersection of events \(A\) and \(B\), expressed as \(A \cap B\), signifies the scenario where both events happen at the same time. Imagine flipping a coin and rolling a dice, where \(A\) is getting a 'Heads' and \(B\) is rolling a 'Six'. The intersection \(A \cap B\) occurs when both a 'Heads' and 'Six' arise.Probabilistically, the value \(\mathrm{P}(A \cap B)\) is derived considering the likelihood of both situations aligning. This is crucial because it provides a quantitative measure of the overlap between \(A\) and \(B\).
In problems like the one given, where \(\mathrm{P}(A \cap B) = \frac{1}{9}\), this value is essential for correctly calculating the probability of the union of events, as it directly impacts the result by adjusting the compounded likelihood found from summing \(\mathrm{P}(A)\) and \(\mathrm{P}(B)\).