Question

A positive integer from one to six is to be chosen by casting a die. Thus the elements \(c\) of the sample space \(\mathcal{C}\) are \(1,2,3,4,5,6 .\) Suppose \(C_{1}=\\{1,2,3,4\\}\) and \(C_{2}=\\{3,4,5,6\\} .\) If the probability set function \(P\) assigns a probability of \(\frac{1}{6}\) to each of the elements of \(\mathcal{C}\), compute \(P\left(C_{1}\right), P\left(C_{2}\right), P\left(C_{1} \cap C_{2}\right)\), and \(P\left(C_{1} \cup C_{2}\right)\).

Short answer

The probability of event \(C_{1}\) is \(\frac{2}{3}\), for event \(C_{2}\) is \(\frac{2}{3}\), for the intersection of \(C_{1}\) and \(C_{2}\) is \(\frac{1}{3}\), and for the union of \(C_{1}\) and \(C_{2}\) is \(1\).

Step-by-step solution

Determine the Probability of Event \(C_{1}\)

The event \(C_{1}\) contains four outcomes: \(1, 2, 3\), and \(4\). Since each outcome has a \(\frac{1}{6}\) probability, the probability of event \(C_{1}\) is \(4\times \frac{1}{6} = \frac{2}{3}\).

Determine the Probability of Event \(C_{2}\)

Similarly, the event \(C_{2}\) contains four outcomes: \(3, 4, 5\), and \(6\). As in step 1, using the probability of each outcome, we find the probability of event \(C_{2}\) is also \(4\times \frac{1}{6} = \frac{2}{3}\).

Determine the Probability of the Intersection of \(C_{1}\) and \(C_{2}\)

The intersection of \(C_{1}\) and \(C_{2}\), written as \(C_{1} \cap C_{2}\), contains the outcomes that are common to both \(C_{1}\) and \(C_{2}\), which are \(3\) and \(4\). Therefore, applying the rule of product in probability, the probability of the intersection of \(C_{1}\) and \(C_{2}\) is \(2\times \frac{1}{6} = \frac{1}{3}\).

Determine the Probability of the Union of \(C_{1}\) and \(C_{2}\)

The union of \(C_{1}\) and \(C_{2}\), written as \(C_{1} \cup C_{2}\), contains all the outcomes that are in either \(C_{1}\), \(C_{2}\), or both. Since the sample space includes numbers from \(1\) to \(6\), the union of \(C_{1}\) and \(C_{2}\) is the same as the sample space. Therefore, the sum of all six equally likely outcomes gives the probability of the union of \(C_{1}\) and \(C_{2}\), which is \(6\times \frac{1}{6}=1\).

Key concepts

Sample Space in Probability Theory

In probability theory, the sample space, often denoted as \( \mathcal{S} \) or \( \Omega \) is a fundamental concept. It's the set that includes all the possible outcomes of a random experiment. For example, if you roll a six-sided die, the sample space is \( \mathcal{C} = \{1,2,3,4,5,6\} \) as these are all possible outcomes. Each outcome in the sample space is equally likely if the die is fair.

The size of a sample space is essential since it affects the computation of probabilities for events. In our example, since there are six outcomes and the die is fair, each outcome's probability is \( \frac{1}{6} \). Understanding the sample space is crucial before you can delve into more complicated concepts in probability.

Probability Set Function

The probability set function, usually denoted as \( P \) is a way to assign a probability to events in a sample space. It maps subsets of the sample space to numbers between \( 0 \) and \( 1 \)—where \( 0 \) signifies the event is impossible, and \( 1 \) indicates it is certain.

In our die example, the function \( P \) assigns a probability of \( \frac{1}{6} \) to each of the elements in the sample space. This means that for any event \( E \) consisting of outcomes from the sample space, the probability \( P(E) \) is calculated by summing the probabilities of each outcome within that event.

For instance, for an event \( C_1 = \{1,2,3,4\} \) with four outcomes, \( P(C_1) \) would be the sum of probabilities of outcomes \( 1 \) through \( 4 \)— giving us \( P(C_1) = 4 \times \frac{1}{6} = \frac{2}{3} \) as seen in the solution.

Intersection of Events

The intersection of events, denoted as \( A \cap B \) where \( A \) and \( B \) are events, refers to a new event that includes all the outcomes common to both \( A \) and \( B \) only. It is the equivalent of the 'and' condition in probability.

Considering the previous events \( C_1 \) and \( C_2 \) from our die example, the intersection \( C_1 \cap C_2 \) would consist of outcomes that appear in both sets— which are \( 3 \) and \( 4 \) only.

The probability of the intersection \( P(C_1 \cap C_2) \) is then the sum of the probabilities of these common outcomes. Since \( 3 \) and \( 4 \) each have a probability of \( \frac{1}{6} \), \( P(C_1 \cap C_2) \) is \( 2 \times \frac{1}{6} = \frac{1}{3} \) as previously calculated in the solutions.

Union of Events

The union of events represents the combination of all outcomes from multiple events, denoted as \( A \cup B \) where \( A \) and \( B \) are two events. It encompasses any outcome that is in either \( A \) or \( B \) or in both, aligning with the 'or' condition in probability.

For instance, if you look at events \( C_1 \) and \( C_2 \) like in our die-casting example, the union \( C_1 \cup C_2 \) would include all the numbers from \( 1 \) to \( 6 \) since these are the numbers in at least one of the sets. Here, the union comprises the entire sample space, indicating that one of these outcomes will certainly occur, which gives us \( P(C_1 \cup C_2) = 1 \)—matching the sample space's total probability.

Understanding the union of events is crucial in scenarios where multiple outcomes lead to a successful or desired result, as it helps to calculate the overall chances of success.