Q.A die is thrown. If $E$ is the event 'the number appearing is a multiple of 3' and $F$ be the event 'the number appearing is even' then find whether $E$ and $F$ are independent?

Concept: Event Independence — Two events $E$ and $F$ are independent if $P(E \cap F) = P(E) \cdot P(F)$.

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

$E = \{3,6\}$, so $P(E) = \frac{2}{6} = \frac{1}{3}$.

$F = \{2,4,6\}$, so $P(F) = \frac{3}{6} = \frac{1}{2}$.

Step 2: $E \cap F = \{6\}$, so $P(E \cap F) = \frac{1}{6}$.

Step 3: Check product: $P(E) \cdot P(F) = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}$.

Since $P(E \cap F) = \frac{1}{6} = P(E) \cdot P(F)$, the events satisfy the independence condition.

The core idea behind independence is simple: one event happening should not change the probability of the other. For a fair die, each face (1 through 6) is equally likely. So we can check this directly by comparing the product of individual probabilities with the probability of both happening together.

Let’s break it down.

1. Define the sample space.

   A single die throw gives $S = \{1, 2, 3, 4, 5, 6\}$, with each outcome having probability $\frac{1}{6}$.

2. Identify the events.

   • $E$: number is a multiple of 3. Multiples of 3 in 1–6 are 3 and 6. So $E = \{3, 6\}$.
   • $F$: number is even. Even numbers are 2, 4, 6. So $F = \{2, 4, 6\}$.

3. Compute individual probabilities.

   • $P(E) = \frac{|E|}{|S|} = \frac{2}{6} = \frac{1}{3}$.
   • $P(F) = \frac{|F|}{|S|} = \frac{3}{6} = \frac{1}{2}$.

4. Find the intersection $E \cap F$.

   The common outcomes: 6 is both a multiple of 3 and even. So $E \cap F = \{6\}$.

   Hence $P(E \cap F) = \frac{1}{6}$.

5. Check the independence condition.

   For independence, we need $P(E \cap F) = P(E) \cdot P(F)$.

   Compute the product:

This matches $P(E \cap F) = \frac{1}{6}$ exactly.

6. Interpret the result. Since the equality holds, $E$ and $F$ are independent events. Knowing the number is even does not change the chance that it is a multiple of 3, and vice versa.

Method: Testing two events for independence

Use this whenever you must decide whether two events are independent — the answer is a single equality check, never a guess based on whether they overlap.

Steps

Step 1: Compute the three probabilities you need.

Find $P(E)$, $P(F)$ and the joint $P(E\cap F)$ by listing the sample space and the relevant subsets.

Step 2: Apply the product test.

Events are independent exactly when

This symmetric form needs no non-zero condition and treats both events alike.

Step 3: Compare and conclude.

If the joint probability equals the product, the events are independent; if not, they are dependent. Sharing a common outcome does not by itself make events dependent — only the numerical equality decides. (You may double-check with $P(E\mid F)=P(E)$, which is equivalent.)

Common Mistakes

Mistake 1: Declaring the events dependent just because they share the outcome $6$.

Why it's wrong: overlap does not decide independence — only the equality $P(E\cap F)=P(E)P(F)$ does. Correct approach: check $P(E\cap F)=\tfrac16$ against $P(E)P(F)=\tfrac13\cdot\tfrac12=\tfrac16$; they match, so the events are independent.

Mistake 2: Confusing independent with mutually exclusive.

Why it's wrong: mutually exclusive events cannot occur together, whereas these can (both happen at $6$). Correct approach: use the product test, not disjointness, to judge independence.