A die is thrown. Let A be the event that the number obtained is greater than 3. Let B be the event t

Q: Wait, why is the number 4 in both sets? Does that change the probability?

That is a fantastic observation! The number 4 is indeed in both sets because it satisfies both conditions: it is greater than 3 and less than 5. In set theory, this is called the intersection, $A \cap B = \{4\}$. When we calculate the union $A \cup B$, we are looking for the set of all elements that belong to either A or B. Because we are dealing with sets, we only list the number 4 once. It doesn't 'double count' the probability; it just means 4 is a member of the union.

Q: Why is the final probability 1? Does that mean the die is rigged?

Not at all! A probability of 1 simply means that the event is a 'certain event.' In this specific problem, every possible outcome of a standard die roll (1, 2, 3, 4, 5, or 6) satisfies either the condition of being greater than 3 or less than 5. Since every outcome is covered, it is guaranteed to happen, hence the probability is 1.

Q: Can I use the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ instead of listing the sets?

Absolutely! That is the Inclusion-Exclusion Principle, and it is a powerful tool. $P(A) = \frac{3}{6}$, $P(B) = \frac{4}{6}$, and $P(A \cap B) = \frac{1}{6}$. Plugging these in: $\frac{3}{6} + \frac{4}{6} - \frac{1}{6} = \frac{6}{6} = 1$. Both methods lead to the same beautiful result.

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

Options:

(A) 3/5

(B) 0

(C) 1

(D) 2/5

Sigma Percentile · Accepted Answer

### The Dance of Chance: Understanding Probability Through a Simple Die Roll

Imagine you are holding a standard, six-sided die. It feels solid, balanced, and fair. When you roll it, you are participating in one of the most fundamental experiments in probability theory.

Today, we are going to explore the beauty of this experiment by analyzing two specific events and their union. Let us embark on this journey together.

### The Universe of Possibilities: Defining the Sample Space

Before we dive into the conditions, we must define our universe. In probability, we call this the sample space, denoted by $S$.

For a standard die, the possible outcomes are simple: $S = \{1, 2, 3, 4, 5, 6\}$. The total number of outcomes, which we denote as $n(S)$, is $6$.

This is our foundation. Every probability we calculate will be relative to this set of six possibilities.

### The Quest for Events: Dissecting A and B

Now, let us define our events. Event $A$ is defined as the outcome being strictly greater than $3$.

On our number line, we look at our sample space and pick out the numbers that satisfy $x > 3$. These are $4, 5,$ and $6$.

So, we have $A = \{4, 5, 6\}$, and the number of favorable outcomes is $n(A) = 3$.

Next, we have event $B$, defined as the outcome being strictly less than $5$. Looking at our sample space again, we pick out the numbers that satisfy $x < 5$.

These are $1, 2, 3,$ and $4$. Thus, $B = \{1, 2, 3, 4\}$, and the number of favorable outcomes is $n(B) = 4$.

### The Art of Union: Merging Worlds

The question asks for $P(A \cup B)$. In set theory, the union $A \cup B$ is the set of all elements that are in $A$, or in $B$, or in both.

Think of it as merging the two sets into one, while being careful not to list duplicates. When we combine $A = \{4, 5, 6\}$ and $B = \{1, 2, 3, 4\}$, we get the set $A \cup B = \{1, 2, 3, 4, 5, 6\}$.

Notice something profound here? The union of these two events is exactly the same as our original sample space $S$. Every single outcome of the die roll is included in this union.

### The Final Calculation: The Certainty of One

Now, we apply the fundamental probability formula:

$$P(E) = \frac{n(E)}{n(S)}$$

For our event $A \cup B$, we have $n(A \cup B) = 6$. Since $n(S) = 6$, the probability is:

$$P(A \cup B) = \frac{6}{6} = 1$$

A probability of $1$ is a special result. It signifies a **certain event**. It means that no matter what number you roll, the condition will be satisfied.

Whether you roll a $1, 2, 3, 4, 5,$ or $6$, it will always be either greater than $3$ or less than $5$. There is no escape from this certainty!

### Beyond the Die: Why This Matters

This problem might seem simple, but it teaches us a vital lesson about the structure of probability. It shows us how events can overlap and how they can collectively cover the entire sample space.

Whether you are analyzing complex quantum systems or simple dice, the logic remains the same: define your space, identify your events, and understand their relationships.

Keep practicing, keep questioning, and most importantly, keep enjoying the elegance of mathematics.

Animated Solution for Mathematics - Probability: 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

Select Answer:

Visualized Solution

The Sample Space $S$

Defining Event $A$

Elements of Event $A$

Defining Event $B$

Elements of Event $B$

Understanding the Union $A \cup B$

Constructing the Union Set

Counting the Union Elements

Applying the Probability Formula

Final Calculation

Conclusion: The Certain Event

The Sigma Insight: Classical Definition of Probability

The Dance of Chance

The Universe of Possibilities

The Quest for Events

The Art of Union

The Final Calculation

Beyond the Die

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