Problem 163 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

A die is thrown. Let A be the event that the number obtained is greater than \(3 .\) Let \(\mathrm{B}\) 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

Answer

The probability \(P(A \cup B)\) is \(1\).

Step 1: Determine the sample space

When a die is thrown, there are 6 possible outcomes: \(1, 2, 3, 4, 5,\) and \(6\). Therefore, the sample space \(S\) is \(\{1, 2, 3, 4, 5, 6\}\).

Step 2: Identify Event A

Event \(A\) is defined as obtaining a number greater than \(3\). Therefore, \(A = \{4, 5, 6\}\).

Step 3: Identify Event B

Event \(B\) is defined as obtaining a number less than \(5\). Therefore, \(B = \{1, 2, 3, 4\}\).

Step 4: Find A ∪ B

The union of two events \(A\) and \(B\), denoted \(A \cup B\), includes all outcomes that are in either \(A\) or \(B\) or both. Therefore, \(A \cup B = \{1, 2, 3, 4, 5, 6\}\).

Step 5: Calculate P(A ∪ B)

Since \(A \cup B\) represents the entire sample space \(S\), the probability \(P(A \cup B) = \frac{|A \cup B|}{|S|} = \frac{6}{6} = 1\).

Key Concepts

Sample Space

In probability theory, the concept of a **sample space** is fundamental. It is the set of all possible outcomes in a given experiment. When dealing with a six-sided die, each side represents a possible outcome when the die is rolled. This means the sample space, denoted as \( S \), is \( \{1, 2, 3, 4, 5, 6\} \).
You can think of the sample space as the universe of all events for this particular experiment.

• If an event occurs, it involves one or more outcomes from this sample space.
• The total number of outcomes determines the size of the sample space.

Understanding the sample space is crucial because it forms the basis for calculating probabilities of events within the space.

Union of Events

The **union of events** combines outcomes from two or more events to show which outcomes satisfy at least one of them. In mathematical terms, the union of two events \( A \) and \( B \) is written as \( A \cup B \).
For our example:

• Event \( A \) is rolling a number greater than \( 3 \), so \( A = \{4, 5, 6\} \).
• Event \( B \) is rolling a number less than \( 5 \), so \( B = \{1, 2, 3, 4\} \).

When finding \( A \cup B \), we include all unique outcomes from both events.
This leads to \( A \cup B = \{1, 2, 3, 4, 5, 6\} \), which covers all possible outcomes when the die is rolled. Therefore, \( A \cup B \) effectively includes the entire sample space.

Probability of Events

The **probability of events** measures the likelihood that any given event occurs. It is calculated using the formula:\[P(Event) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}\]
In our example, we want to find the probability of the union of events \( A \) and \( B \), which we have established as \( A \cup B = \{1, 2, 3, 4, 5, 6\} \).

• The number of favorable outcomes in \( A \cup B \) is \( 6 \), since it includes all outcomes in the sample space.
• The total number of possible outcomes in the sample space \( S \) is \( 6 \).

Thus, the probability \( P(A \cup B) = \frac{|A \cup B|}{|S|} = \frac{6}{6} = 1 \).
This probability of \( 1 \) confirms that the union of \( A \) and \( B \) involves all scenarios, meaning one of the events (either \( A \) or \( B \)) will always occur when rolling the die.