Q.An unbiased die is thrown twice. Let the event $A$ be 'odd number on the first throw' and $B$ the event 'odd number on the second throw'. Check the independence of the events $A$ and $B$.

Concept: Event Independence — Two events are independent if $P(A \cap B) = P(A) \cdot P(B)$.

Step 1: For a fair die, each throw has 3 odd numbers out of 6.

$P(A) = \frac{3}{6} = \frac{1}{2}$, and $P(B) = \frac{1}{2}$.

Step 2: The event $A \cap B$ means odd on both throws. There are $3 \times 3 = 9$ favourable outcomes out of $6 \times 6 = 36$ total.

$P(A \cap B) = \frac{9}{36} = \frac{1}{4}$.

Step 3: Check the product: $P(A) \cdot P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.

Since $P(A \cap B) = P(A) \cdot P(B)$, the events are independent.

Why Independence Matters Here

When we say two events are independent, we mean that knowing whether one event happened gives you no information about whether the other event happened. In the context of throwing a die twice, the result of the first throw has absolutely no effect on the second throw — the die has no memory. So intuitively, A and B should be independent.

But intuition isn't proof. We need to check the mathematical definition: two events A and B are independent if and only if $P(A \cap B) = P(A) \cdot P(B)$.

Let's work through it step by step.

1. Find $P(A)$ — probability of odd number on the first throw

A fair die has six faces: 1, 2, 3, 4, 5, 6. The odd numbers are 1, 3, 5 — three outcomes out of six.

2. Find $P(B)$ — probability of odd number on the second throw

Exactly the same reasoning applies to the second throw. The second throw is independent of the first by the nature of the experiment.

3. Find $P(A \cap B)$ — probability that both throws show odd numbers

The sample space for two throws has $6 \times 6 = 36$ equally likely outcomes. For both throws to be odd, the first throw must be one of {1, 3, 5} and the second must also be one of {1, 3, 5}. That gives $3 \times 3 = 9$ favourable outcomes.

4. Check the independence condition

Now compute $P(A) \cdot P(B)$:

Compare with $P(A \cap B) = \frac{1}{4}$. They are equal.

5. Conclusion

Since $P(A \cap B) = P(A) \cdot P(B)$, the events A and B satisfy the definition of independence.

Method: Testing independence of events tied to different trials

Use this when two events refer to different repetitions of an experiment (first throw vs second throw) and you must verify independence formally rather than assert it.

Steps

Step 1: Find each marginal probability.

Compute $P(A)$ and $P(B)$ from a single trial's outcomes (e.g. "odd on a throw" is $\tfrac{3}{6}=\tfrac12$).

Step 2: Find the joint probability from the full two-trial space.

Count favourable ordered outcomes over the total ($6\times6=36$ for two dice): both-odd gives $3\times3=9$, so $P(A\cap B)=\tfrac{9}{36}=\tfrac14$. Get this from the sample space, not by assuming independence.

Step 3: Check the product rule.

Verify $P(A\cap B) = P(A)\,P(B)$. Deriving the joint by multiplying already assumes the result, so for a genuine proof you must obtain $P(A\cap B)$ independently (by counting) and then confirm the equality holds.

Common Mistakes

Mistake 1: Assuming independence and multiplying $\tfrac12\times\tfrac12$ as the "proof."

Why it's wrong: writing $P(A\cap B)=P(A)P(B)$ from the start assumes exactly what must be proven. Correct approach: obtain $P(A\cap B)=\tfrac{9}{36}=\tfrac14$ by counting the $36$-outcome space independently, then confirm it equals the product.

Mistake 2: Thinking a non-empty intersection implies dependence.

Why it's wrong: independence is a numerical condition, not about whether the events can co-occur. Correct approach: rely solely on $P(A\cap B)=P(A)P(B)$.