Independent events

Toss a coin and roll a die at the same time. Does knowing the coin landed heads tell you anything about what the die will show? No — the two outcomes have nothing to do with each other, and probability theory gives this everyday intuition a precise, checkable definition. Events $A$ and $B$ are independent exactly when $P(A\cap B) = P(A)\,P(B)$ : a single equation you can test by calculation, not by guessing from how "related" two events sound. This idea sits right after the multiplication theorem in the CBSE Class 12 Probability chapter, and it is one of the most confused topics in the whole unit — students routinely mix up independence with mutual exclusivity, two properties that are almost opposites. This article nails down the definition, extends it to three events, derives a useful consequence for complements, and works through fully-solved dice-and-coin examples so the test becomes second nature.

The core insight

Two events are independent when $P(A\cap B) = P(A)P(B)$ , so knowing one occurred changes nothing about the probability of the other.

The core idea

Definition. Events $A$ and $B$ , with $P(A) > 0$ and $P(B) > 0$ , are called independent if their joint probability factorises into the product of the individual probabilities:

P(A\cap B) = P(A)\,P(B).

where: $P(A\cap B)$ is the probability that both $A$ and $B$ occur, and $P(A)$ , $P(B)$ are their individual probabilities.

Equivalent conditional form. By the multiplication theorem, $P(A\cap B) = P(A)\,P(B\mid A)$ in general. Substituting the independence condition shows $P(B\mid A) = P(B)$ and, symmetrically, $P(A\mid B) = P(A)$ . So independence can equally be stated as: conditioning on one event leaves the probability of the other unchanged.

Three events — mutual independence. Events $A$ , $B$ , $C$ are mutually independent if all four conditions hold together:

P(A\cap B) = P(A)P(B), \quad P(B\cap C) = P(B)P(C), \quad P(A\cap C) = P(A)P(C),

P(A\cap B\cap C) = P(A)\,P(B)\,P(C).

where: the first three are the pairwise conditions and the fourth is the joint condition for all three together. If only the first three hold but not the fourth, the events are pairwise independent but not mutually independent — pairwise checks alone never guarantee mutual independence, so all four must be verified separately.

Independence versus mutual exclusivity. These two properties are near-opposites, and confusing them is the single most common mistake in this chapter. Mutually exclusive events satisfy $A\cap B = \varnothing$ , so $P(A\cap B) = 0$ . If both events also have positive probability, then $P(A)P(B) > 0 \ne 0 = P(A\cap B)$ , so mutually exclusive events with positive probability can never be independent. Mutual exclusivity says the events cannot happen together at all; independence says whether one happens has no bearing on the other. They describe entirely different relationships between events.

Figure

Two Venn diagrams contrasting mutually exclusive circles that do not overlap with independent circles that overlap in proportion to their sizes

Derivation

Derivation AIndependence of the complements: if A and B are independent, so are A and B'

The goal is to show that independence is a robust structural property, not a fragile numerical coincidence tied to $B$ alone: prove that if $A$ and $B$ satisfy $P(A\cap B) = P(A)P(B)$ , then $A$ and the complement $B'$ (the event " $B$ does not occur") also satisfy the independence condition $P(A\cap B') = P(A)P(B')$ , using only the addition theorem and basic set algebra already known from earlier in the chapter.

Step 1. Split $A$ using $B$ and its complement. Every outcome in $A$ either also lies in $B$ or does not, so $A$ is the disjoint union of $A\cap B$ and $A\cap B'$ : $A = (A\cap B) \cup (A\cap B')$ , with $(A\cap B)$ and $(A\cap B')$ mutually exclusive.

Step 2. Apply the addition theorem for mutually exclusive events. Since the two pieces are disjoint, $P(A) = P(A\cap B) + P(A\cap B')$ . Rearranging gives $P(A\cap B') = P(A) - P(A\cap B)$ .

Step 3. Substitute the independence assumption for $A$ and $B$ . We are given $P(A\cap B) = P(A)P(B)$ , so $P(A\cap B') = P(A) - P(A)P(B) = P(A)\big(1 - P(B)\big)$ .

Step 4. Recognise $1 - P(B)$ as $P(B')$ . By the complement rule, $P(B') = 1 - P(B)$ , so $P(A\cap B') = P(A)\,P(B')$ — exactly the independence condition for $A$ and $B'$ . The same argument, run with $A$ and $A'$ instead, shows $A'$ and $B$ are independent, and repeating it once more shows $A'$ and $B'$ are independent too. Independence of two events therefore automatically carries over to their complements — a fact worth knowing rather than re-deriving under exam pressure.

Worked example

Two dice are thrown together. Let $A$ = "the sum of the numbers is 7" and $B$ = "the first die shows a 4." Using the result just derived, check whether $A$ and $B'$ (first die does not show a 4) are independent, and find $P(A\cap B')$ .

Step 1. Recall the base facts from this pair of events. There are 36 equally likely outcomes. $A$ = sum is 7 has 6 outcomes, so $P(A) = \dfrac{6}{36} = \dfrac16$ . $B$ = first die shows 4 has 6 outcomes, so $P(B) = \dfrac16$ . Only $(4,3)$ lies in both, so $P(A\cap B) = \dfrac{1}{36}$ , and since $P(A)P(B) = \dfrac16\times\dfrac16 = \dfrac{1}{36}$ , $A$ and $B$ were already confirmed independent.

Step 2. Find $P(B')$ . Since $B'$ is "first die is not 4," $P(B') = 1 - P(B) = 1 - \dfrac16 = \dfrac56$ .

Step 3. Apply the derived result. Because $A$ and $B$ are independent, the derivation guarantees $A$ and $B'$ are independent too, so $P(A\cap B') = P(A)\,P(B') = \dfrac16 \times \dfrac56 = \dfrac{5}{36}$ .

Step 4. Verify by direct counting. $A\cap B'$ means the sum is 7 and the first die is not 4. Of the six sum-7 outcomes $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ , only $(4,3)$ has first die equal to 4, so five of the six satisfy $A\cap B'$ : that is $\dfrac{5}{36}$ of all 36 outcomes. This matches Step 3 exactly, confirming $P(A\cap B') = \dfrac{5}{36}$ and that $A$ and $B'$ are independent.

Competitive stretch

Using the same $A$ and $B$ , find $P(A\cup B)$ . Since $A$ and $B$ are independent, apply the addition theorem with the independence value of $P(A\cap B)$ : $P(A\cup B) = P(A) + P(B) - P(A\cap B) = \dfrac16 + \dfrac16 - \dfrac{1}{36} = \dfrac{6}{36}+\dfrac{6}{36}-\dfrac{1}{36} = \dfrac{11}{36}$ . Independence never removes the need for the addition theorem when computing a union — it only simplifies the intersection term inside it.

Common traps

Treating mutually exclusive as independent, or the reverse. If $A$ and $B$ are mutually exclusive with positive probabilities, they are never independent; always check $P(A\cap B) = 0 \ne P(A)P(B)$ in that case rather than assuming a relationship either way.
Assuming pairwise independence for three events implies mutual independence. All four conditions — three pairwise products and the triple product — must be checked separately; a set of events can pass all three pairwise tests and still fail the fourth.
Calling events independent just because they "sound unrelated." Independence is the numerical condition $P(A\cap B) = P(A)P(B)$ , verified by calculation, not a judgement call based on how the events are worded.
Forgetting that independence carries to complements only under the derived structure. $P(A\cap B') = P(A)P(B')$ is a consequence of $A, B$ being independent — it is not a separate assumption to invent, but it also is not automatic without that starting independence.

Ask the tutor: Still mixing up independence with mutual exclusivity, or want to practise the three-event mutual independence check? Ask the AI tutor for more worked examples targeting exactly that distinction.

Why is it wrong to say two dice showing different numbers are independent just because dice don't affect each other physically?

Physical separateness is a good hint but not a proof. "Independent" in probability has one precise test: compute $P(A\cap B)$ and $P(A)P(B)$ separately and check they are equal. Some physically-separate-sounding events on the same pair of dice can still fail this test, especially when both events are defined using the same die or the same sum. Always run the numbers rather than trusting intuition about physical causation.

Can an event be independent of itself?

Only in a degenerate case. Setting $B = A$ in the definition gives $P(A\cap A) = P(A)P(A)$ , i.e. $P(A) = P(A)^2$ , which only holds when $P(A) = 0$ or $P(A) = 1$ . So a genuine event with probability strictly between 0 and 1 is never independent of itself — makes sense, since knowing $A$ occurred tells you with certainty that $A$ occurred.

For three events, why isn't checking all three pairs enough to conclude mutual independence?

Because the pairwise conditions only constrain how each pair of events overlaps two at a time; they say nothing about how all three overlap together. It is possible to construct events where every pair multiplies correctly but the triple intersection $P(A\cap B\cap C)$ does not equal $P(A)P(B)P(C)$ — a classic counterexample uses events built from two coin tosses. CBSE explicitly tests this distinction, so always verify the fourth condition separately rather than assuming it follows.