Q.A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let $A$ be the event 'the number is even,' and $B$ be the event, 'the number is red'. Are $A$ and $B$ independent?

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

Step 1: List the sample space.

$S = \{1,2,3,4,5,6\}$, each equally likely with probability $\frac{1}{6}$.

Step 2: Compute probabilities.

• $A$ (even number): $\{2,4,6\} \Rightarrow P(A) = \frac{3}{6} = \frac{1}{2}$.
• $B$ (red number): $\{1,2,3\} \Rightarrow P(B) = \frac{3}{6} = \frac{1}{2}$.
• $A \cap B$ (even and red): only $\{2\} \Rightarrow P(A \cap B) = \frac{1}{6}$. …

The core idea behind independence is simple: two events are independent if knowing that one has happened gives you no information about whether the other has happened. In probability terms, this means the probability of both occurring together is exactly the product of their individual probabilities.

Let’s check whether that holds here.

1. Define the sample space.

   A fair die has six equally likely outcomes: $\{1, 2, 3, 4, 5, 6\}$. Each outcome has probability $\frac{1}{6}$.

2. Identify event $A$ — 'the number is even'.

   The even numbers on a die are $2, 4, 6$. So $A = \{2, 4, 6\}$.

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

3. Identify event $B$ — 'the number is red'.

   The problem states that numbers 1, 2, 3 are red, and 4, 5, 6 are green. So $B = \{1, 2, 3\}$.

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

4. Find the intersection $A \cap B$.

   This is the set of outcomes that are both even and red.

   Even numbers: $\{2, 4, 6\}$. Red numbers: $\{1, 2, 3\}$.

   The only common number is $2$. So $A \cap B = \{2\}$.

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

5. Check the independence condition.

   For independence, we need $P(A \cap B) = P(A) \cdot P(B)$. …

Method: Independence Check by Listing the Event Sets

Use this when events are described in words on one experiment ("even", "red", etc.) and you must decide whether they are independent.

Steps

Step 1: Translate each event into an outcome set

Write the sample space, then list the outcomes making up $A$ and $B$ explicitly. From the set sizes read off $P(A)$ and $P(B)$.

Step 2: Take the intersection set

Intersect the two lists to get $A\cap B$ — the outcomes that satisfy both conditions at once — and count them to get $P(A\cap B)$. Getting the overlap set right is where most errors happen. …

Common Mistakes

Mistake 1: Assuming independence because $P(A)=P(B)=\frac12$.

Equal individual probabilities say nothing about independence. Why it's wrong: independence depends on the overlap, not the marginals. Correct approach: compare $P(A\cap B)$ with $P(A)P(B)=\frac14$; here $P(A\cap B)=\frac16\neq\frac14$, so not independent.

Mistake 2: Miscounting $A\cap B$. …