Problem 25
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 [2008] (a) \(\frac{3}{5}\) (b) 0 (c) 1 (d) \(\frac{2}{5}\)
Step-by-Step Solution
Verified Answer
The probability \(P(A \cup B)\) is 1.
1Step 1: Identify the Sample Space
When a single die is thrown, the possible outcomes (sample space) are \ \( \{1, 2, 3, 4, 5, 6\} \). This represents the 6 possible faces of a die.
2Step 2: Define Event A
Event \( A \) is defined as rolling a number greater than 3. The numbers that satisfy this condition are \( \{4, 5, 6\} \).
3Step 3: Define Event B
Event \( B \) is defined as rolling a number less than 5. The numbers that satisfy this condition are \( \{1, 2, 3, 4\} \).
4Step 4: Find \(A \cup B\)
The event \( A \cup B \) is the union of events \( A \) and \( B \), meaning all outcomes that are in either \( A \) or \( B \) or both. Thus, \( A \cup B = \{1, 2, 3, 4, 5, 6\} \).
5Step 5: Calculate Probability of \(A \cup B\)
The probability of the union of events \( A \) and \( B \) is \( P(A \cup B) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{6} = 1 \).
Key Concepts
Sample SpaceUnion of EventsEvent ProbabilityDiscrete Mathematics
Sample Space
When dealing with probabilities, one of the first things to understand is the concept of a sample space. The sample space is the set of all possible outcomes of a random experiment. For example, when we throw a single die, there are 6 possible outcomes since a standard die has six faces numbered from 1 to 6.
- Each outcome is equally likely when the die is fair.
- The sample space is often denoted by the letter "S".
Union of Events
The union of two events in probability theory refers to the set of outcomes that are in either one event or the other, or in both. It combines all the outcomes from the events into one set. In our die example, we have events \( A \) and \( B \): - Event \( A \) (rolling a number greater than 3): \( \{4, 5, 6\} \)- Event \( B \) (rolling a number less than 5): \( \{1, 2, 3, 4\} \)The union of these two events, denoted as \( A \cup B \), includes any number found in either event, without duplication. This gives us \[ A \cup B = \{1, 2, 3, 4, 5, 6\} \], which in this case is the entire sample space. To determine the probability of the union of these events, we look at all the distinct outcomes included in both events.
Event Probability
Probability quantifies the likelihood of an event occurring. It is calculated by dividing the number of favorable outcomes by the total number of possible outcomes in the sample space.For the union of events \( A \) and \( B \) in our die example, the event probability of \( A \cup B \) is calculated as follows:- Total number of outcomes in the sample space: 6- Number of favorable outcomes for \( A \cup B \): 6The probability formula \[P(A \cup B) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\]Substituting the numbers, \[P(A \cup B) = \frac{6}{6} = 1\]This means that the event \( A \cup B \) is certain to happen in this case, showing why understanding probabilities is crucial for predicting outcomes.
Discrete Mathematics
Discrete mathematics is a branch of mathematics that deals with distinct and countable objects. It plays a crucial role in computer science, information theory, and combinatorics, among other fields. Probability is an important part of discrete mathematics, especially when dealing with finite sample spaces as demonstrated with our die example.
- Problems in discrete mathematics often involve counting methods such as permutations and combinations.
- It helps in solving problems involving integers or any objects that have a distinct separation.
- Understanding how discrete sets of data function is key to concepts in probability and the broader field of discrete mathematics.
Other exercises in this chapter
Problem 23
If the events \(\mathrm{A}\) and \(\mathrm{B}\) are mutually exclusive events such that \(\mathrm{P}(\mathrm{A})=\frac{3 x+1}{3}\) and \(\mathrm{P}(\mathrm{B})=
View solution Problem 24
Let \(X\) and \(Y\) are two events such that \(P(X \cup Y)=P(X \cap Y)\) Statement 1: \(P\left(X \cap Y^{\prime}\right)=P\left(X^{\prime} \cap Y\right)=0\) Stat
View solution Problem 26
Events \(A, B, C\) are mutually exclusive events such that \(P(A)=\frac{3 x+1}{3}, P(B)=\frac{1-x}{4}\) and \(P(C)=\frac{1-2 x}{2}\) The set of possible values
View solution Problem 27
\(A\) and \(B\) are events such that \(P(A \cup B)=3 / 4, P(A \cap B)=1 / 4\), \(P(\bar{A})=2 / 3\) then \(P(\bar{A} \cap B)\) is \(\quad\) [2002] (a) \(5 / 12\
View solution