Q.A fair coin and an unbiased die are tossed. Let $A$ be the event 'head appears on the coin' and $B$ be the event '3 on the die'. Check whether $A$ and $B$ are independent events or not.

Concept: Conditional Probability / Independence

Two events are independent if $P(A \cap B) = P(A) \cdot P(B)$.

Step 1 – Individual probabilities

For a fair coin: $P(A) = \frac{1}{2}$.

For an unbiased die: $P(B) = \frac{1}{6}$.

Step 2 – Joint probability

Since the coin and die are tossed together, the sample space has $2 \times 6 = 12$ equally likely outcomes. Only one outcome gives both a head and a 3: $(H,3)$.

So $P(A \cap B) = \frac{1}{12}$.

Step 3 – Check the product …

Why this works — the idea of independence

Independence means that knowing whether one event happened gives you no information about whether the other happened. For coin and die tosses, that's intuitively true: the coin doesn't care what the die shows, and vice versa. But we need to check it formally.

The mathematical test is simple: two events $A$ and $B$ are independent if and only if

If this equality holds, they're independent. If it doesn't, they're dependent.

Step-by-step verification

1. Find $P(A)$ — the probability of heads on the coin.

A fair coin has two equally likely outcomes: heads or tails.

2. Find $P(B)$ — the probability of a 3 on the die.

An unbiased die has six equally likely faces: 1 through 6. Only one face shows 3.

3. Find $P(A \cap B)$ — the probability that both happen together.

The coin and die are tossed simultaneously. The sample space has $2 \times 6 = 12$ equally likely outcomes:

Only one outcome — $(H,3)$ — satisfies both "heads on coin" and "3 on die".

$P(A \cap B) = \frac{1}{12}.$ …

Method: Testing Whether Two Events Are Independent

Use this whenever an experiment defines two events and you must decide, by calculation, whether they are independent.

Steps

Step 1: Find the two marginal probabilities

Work out $P(A)$ and $P(B)$ separately from the experiment.

Step 2: Find the joint probability directly

Compute $P(A\cap B)$ from the actual outcomes — list or count the sample-space points that satisfy both events (for a coin-and-die type experiment the joint sample space has $|S_1|\times|S_2|$ equally likely outcomes). Do not assume the product yet; find $P(A\cap B)$ independently. …

Common Mistakes

Mistake 1: Assuming the events must be dependent because the coin and die are tossed together.

Being part of one experiment does not link the outcomes. Why it's wrong: a coin result carries no information about the die. Correct approach: test $P(A\cap B)=P(A)P(B)$; here $\frac{1}{12}=\frac12\cdot\frac16$, so they are independent.

Mistake 2: Mis-sizing the joint sample space. …