Q.Consider the experiment of rolling a die. Let A be the event 'getting a prime number', B be the event 'getting an odd number'. Write the sets representing the events

The key idea is to translate the English descriptions into set operations using the sample space of a die roll.

Step 1: Define the sample space and events.

Sample space: $S = \{1,2,3,4,5,6\}$.

Prime numbers on a die: $A = \{2,3,5\}$.

Odd numbers: $B = \{1,3,5\}$.

Step 2: Apply the set operations.

1. “A or B” means $A \cup B = \{1,2,3,5\}$.
2. “A and B” means $A \cap B = \{3,5\}$. …

The foundation of this problem is understanding how the language of probability maps onto set operations. When we roll a standard six-sided die, the sample space — the set of all possible outcomes — is $S = \{1, 2, 3, 4, 5, 6\}$.

An "event" is just a subset of this sample space. So first, let's write down exactly what A and B contain.

A is "getting a prime number". The prime numbers between 1 and 6 are 2, 3, and 5. So $A = \{2, 3, 5\}$.

B is "getting an odd number". The odd numbers on a die are 1, 3, and 5. So $B = \{1, 3, 5\}$.

Now, each part of the question asks for a different combination of these sets. The trick is to read the English carefully.

1. "A or B" — In everyday language, "or" usually means "either A, or B, or both". In set theory, this is the union of the two sets, written $A \cup B$. It contains every outcome that belongs to A, or to B, or to both.

   • From $A = \{2, 3, 5\}$ and $B = \{1, 3, 5\}$, the union is $\{1, 2, 3, 5\}$.
   • Notice that 3 and 5 appear in both sets, but we list them only once in the union.

2. "A and B" — This means an outcome must satisfy both conditions simultaneously: it must be prime and odd. In set theory, this is the intersection, written $A \cap B$. It contains only the elements common to both sets.

   • The numbers that are in both A and B are 3 and 5. So $A \cap B = \{3, 5\}$.
   • Check: 2 is prime but not odd, so it's out. 1 is odd but not prime, so it's out.

3. "A but not B" — This means outcomes that are in A but are not in B. In set theory, this is the set difference, written $A - B$ or $A \setminus B$. It's like taking A and removing everything that also appears in B.

   • A has $\{2, 3, 5\}$. B has $\{1, 3, 5\}$. Removing the common elements (3 and 5) from A leaves just $\{2\}$.
   • So $A - B = \{2\}$. …