Problem 58 - Applied Statistics and Probability for Engineers

Question

Each of the possible five outcomes of a random experiment is equally likely. The sample space is \(\\{a, b, c, d, e\\} .\) Let \(A\) denote the event \(\\{a, b\\},\) and let \(B\) denote the event \(\\{c, d, e\\} .\) Determine the following: (a) \(P(A)\) (b) \(P(B)\) (c) \(P\left(A^{\prime}\right)\) (d) \(P(A \cup B)\) (e) \(P(A \cap B)\)

Step-by-Step Solution

Answer

(a) \(\frac{2}{5}\); (b) \(\frac{3}{5}\); (c) \(\frac{3}{5}\); (d) 1; (e) 0;

Step 1: Understanding the sample space and probability

The sample space has five equally likely outcomes: \(\{a, b, c, d, e\}\). Since each outcome is equally likely, the probability of any single outcome is \(\frac{1}{5}\).

Step 2: Calculate P(A)

Event \(A\) consists of the outcomes \(\{a, b\}\). Since each outcome has a probability of \(\frac{1}{5}\), the probability of event \(A\) is the sum of the probabilities of its outcomes: \(P(A) = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}\).

Step 3: Calculate P(B)

Event \(B\) consists of the outcomes \(\{c, d, e\}\). The probability of event \(B\) is the sum of the probabilities of its outcomes: \(P(B) = \frac{1}{5} + \frac{1}{5} + \frac{1}{5} = \frac{3}{5}\).

Step 4: Calculate P(A')

\(A'\) represents the complement of event \(A\), meaning the outcomes not in \(A\), which are \(\{c, d, e\}\). Thus, \(P(A') = P(B) = \frac{3}{5}\).

Step 5: Calculate P(A ∪ B)

The event \(A \cup B\) represents the union of \(A\) and \(B\). Since \(A\) and \(B\) cover the entire sample space \(\{a, b, c, d, e\}\), \(P(A \cup B) = 1\).

Step 6: Calculate P(A ∩ B)

The event \(A \cap B\) represents the intersection of \(A\) and \(B\), which are the outcomes common to both events. Since \(A\) and \(B\) share no common outcomes, \(P(A \cap B) = 0\).

Key Concepts

Sample Space

In probability theory, a sample space is the set of all possible outcomes of a random experiment. It's like the complete deck of possibilities for what can happen in your experiment. For example, consider the experiment of tossing a die. The sample space for this experiment is the set \(\{1, 2, 3, 4, 5, 6\}\), representing all possible outcomes when you roll the die.

In our exercise, the sample space is \(\{a, b, c, d, e\}\). Since each of these outcomes is equally likely, they each have a probability of \(\frac{1}{5}\). Understanding the sample space is crucial because it serves as the foundation on which we calculate probabilities of other events.

Complement of an Event

The complement of an event in probability is everything that is not part of the event within the sample space. Essentially, if you have an event, its complement is the opposite of that event happening. For any event \(A\), the complement is denoted as \(A'\). The probability of an event plus the probability of its complement always equals 1.

In our exercise, event \(A\) consists of the outcomes \(\{a, b\}\). Thus, the complement of \(A\) would be the outcomes that are not \(a\) or \(b\), which are \(\{c, d, e\}\). Consequently, \(P(A') = \frac{3}{5}\), as these are the remaining possible outcomes when \(A\) does not occur.

Union of Events

The union of two events in probability is the set of outcomes that are in either one event or the other, or in both. Think of it as combining all possible outcomes from both events. The union is denoted as \(A \cup B\). This concept is important for understanding how to calculate the probability that at least one of several events will occur.

In our problem, the union \(A \cup B\) represents all the outcomes covered by events \(A\) and \(B\). Since \(A\) and \(B\) together encompass the entire sample space \(\{a, b, c, d, e\}\), the probability \(P(A \cup B) = 1\). This indicates that by selecting an outcome at random, you will definitely select one from either \(A\) or \(B\).

Intersection of Events

The intersection of events refers to outcomes that are common to both events. Represented by \(A \cap B\), this concept narrows down to the outcomes that both events share. If the intersection is empty, it usually implies that the events are mutually exclusive, meaning they cannot both happen at the same time.

For our exercise, \(A = \{a, b\}\) and \(B = \{c, d, e\}\). Since there are no shared outcomes between \(A\) and \(B\), \(A \cap B\) is an empty set, and therefore \(P(A \cap B) = 0\). This outcome emphasizes that there is zero probability of both \(A\) and \(B\) happening together in this scenario.