Problem 69
Question
\(A\) die is thrown. Let \(A\) be the event that the number obtained is greater than \(3 .\) Let \(B\) be the event that the number obtained is less than 5 . then \(P(A \cup B)\) is (A) \(3 / 5\) (B) 0 (C) 1 (D) \(2 / 5\)
Step-by-Step Solution
Verified Answer
The probability \( P(A \cup B) \) is 1 (option C).
1Step 1: Identify the Sample Space
Since a die is thrown, the sample space is the set of all possible outcomes: \( \{1, 2, 3, 4, 5, 6\} \). This means there are 6 possible outcomes when a die is thrown.
2Step 2: Define Events A and B
Event \( A \) is defined as the number obtained being greater than 3, i.e., \( A = \{4, 5, 6\} \). Event \( B \) is defined as the number obtained being less than 5, i.e., \( B = \{1, 2, 3, 4\} \).
3Step 3: Find the Union of Events A and B
The union of events \( A \) and \( B \) (\( A \cup B \)) is the set of outcomes that are in either \( A \) or \( B \) or both. Therefore, \( A \cup B = \{1, 2, 3, 4, 5, 6\} \).
4Step 4: Calculate the Probability of A ∪ B
To find \( P(A \cup B) \), count the number of elements in \( A \cup B \) and divide by the total number of possible outcomes (6). Since \( A \cup B \) is the entire sample space \( \{1, 2, 3, 4, 5, 6\} \), there are 6 elements in \( A \cup B \). Thus, \( P(A \cup B) = \frac{6}{6} = 1 \).
Key Concepts
Sample SpaceUnion of EventsEvent ProbabilityDiscrete Mathematics
Sample Space
When dealing with probability, the first step is often identifying the sample space. This is the complete list of all possible outcomes of a particular experiment. In the case of rolling a die, the sample space consists of six potential outcomes: \( \{1, 2, 3, 4, 5, 6\} \). Each number represents a face of the die that might land facing up after a roll.
Understanding the sample space is crucial as it forms the foundation upon which probabilities are calculated. Every outcome has an equal chance of occurring when a fair die is thrown, and each outcome is an important part of making calculations for any events involving the die roll.
Understanding the sample space is crucial as it forms the foundation upon which probabilities are calculated. Every outcome has an equal chance of occurring when a fair die is thrown, and each outcome is an important part of making calculations for any events involving the die roll.
Union of Events
The concept of the union of events in probability helps us to understand scenarios where multiple conditions can lead to favorable results. For events \( A \) and \( B \), the union (denoted as \( A \cup B \)) consists of all outcomes that belong either to event \( A \), to event \( B \), or to both. Essentially, it's a way to combine all outcomes related to the two events, ensuring no possibility is missed.
During our die-rolling example, event \( A \) was defined as numbers greater than 3, which included \( \{4, 5, 6\} \). Event \( B \) was numbers less than 5, covering \( \{1, 2, 3, 4\} \). The union, \( A \cup B \), combined these to give \( \{1, 2, 3, 4, 5, 6\} \), which is the entire sample space. This shows how the union can encompass all possible outcomes when events overlap.
During our die-rolling example, event \( A \) was defined as numbers greater than 3, which included \( \{4, 5, 6\} \). Event \( B \) was numbers less than 5, covering \( \{1, 2, 3, 4\} \). The union, \( A \cup B \), combined these to give \( \{1, 2, 3, 4, 5, 6\} \), which is the entire sample space. This shows how the union can encompass all possible outcomes when events overlap.
Event Probability
The probability of an event gives us a measure of the likelihood that the event will occur. It is calculated by dividing the number of favorable outcomes by the total number of possible outcomes within the sample space.
The exercise involves calculating the probability of the union \( A \cup B \). Since this union is the entire sample space, there are 6 favorable outcomes, as every roll of the die results in success with respect to \( A \cup B \). Thus the probability is \( \frac{6}{6} = 1 \), indicating certainty that one of these will occur. This shows the possible range of probability from 0 (impossible event) to 1 (certain event).
The exercise involves calculating the probability of the union \( A \cup B \). Since this union is the entire sample space, there are 6 favorable outcomes, as every roll of the die results in success with respect to \( A \cup B \). Thus the probability is \( \frac{6}{6} = 1 \), indicating certainty that one of these will occur. This shows the possible range of probability from 0 (impossible event) to 1 (certain event).
Discrete Mathematics
Discrete mathematics involves studying mathematical structures that are fundamentally countable or distinct. It plays a big role in understanding and solving problems where the outcomes are separate and distinct, such as the roll of a die.
This field underlines principles like the sample space and probability calculations, which rely on discrete values. Rolling a die gives a finite, countable set of results—represented as distinct numbers from 1 to 6—in contrast to continuous systems that would include infinite possibilities. Discrete mathematics forms the backbone of probability, statistics, and various algorithms used in computer science, helping model scenarios with clear, defined outcomes.
This field underlines principles like the sample space and probability calculations, which rely on discrete values. Rolling a die gives a finite, countable set of results—represented as distinct numbers from 1 to 6—in contrast to continuous systems that would include infinite possibilities. Discrete mathematics forms the backbone of probability, statistics, and various algorithms used in computer science, helping model scenarios with clear, defined outcomes.
Other exercises in this chapter
Problem 67
A class consists of 80 students, 25 of them are girls. If 10 of the students are rich and 20 of the students are fair complexioned, then the probability of sele
View solution Problem 68
It is given that the events \(A\) and \(B\) are such that \(P(A)=\) \(\frac{1}{4}, P\left(\frac{A}{B}\right)=\frac{1}{2}\) and \(P\left(\frac{B}{A}\right)=\frac
View solution Problem 70
A bag contains \(n+1\) coins. It is known that one of these coins shows heads on both sides, whereas the other coins are fair. One coin is selected at random an
View solution Problem 71
The probabilites of three events \(A, B\) and \(C\) are \(P(A)=0.6, P(B)=0.4\) and \(P(C)=0.5 .\) If \(P(A \cup B)\) \(=0.8, P(A \cap C)=0.3, P(A \cap B \cap C)
View solution