Independent Events

When information changes nothing

Conditional probability measures how the chance of $E$ updates once we learn that $F$ has occurred. Sometimes, though, the update does nothing at all — knowing $F$ leaves the probability of $E$ exactly where it was:

$$P(E|F) = P(E)$$

When this happens, the information "$F$ occurred" is irrelevant to $E$. We then call $E$ and $F$ independent events.

A quick example. Draw one card from a well-shuffled deck of 52. Let $E$ = "the card is a spade" and $F$ = "the card is an ace".

$$P(E) = \frac{13}{52} = \frac{1}{4}, \qquad P(F) = \frac{4}{52} = \frac{1}{13}, \qquad P(E \cap F) = P(\text{ace of spades}) = \frac{1}{52}$$

Now condition:

$$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{1/52}{1/13} = \frac{1}{4} = P(E)$$

Being told "it's an ace" tells you nothing about whether it is a spade — exactly one of the four aces is a spade, just as exactly one-quarter of the whole deck is spades. The events are independent.

The official definition

Substituting $P(E|F) = P(E)$ into the multiplication theorem $P(E \cap F) = P(F)\,P(E|F)$ gives the symmetric, preferred form of the definition.

Definition. Two events $E$ and $F$ are independent if $$P(E \cap F) = P(E) \cdot P(F)$$

This form is better than $P(E|F) = P(E)$ for two reasons: it is symmetric in $E$ and $F$ (no need to decide which event is the "condition"), and it makes sense even when $P(E)$ or $P(F)$ is $0$ (a conditional like $P(E|F)$ needs $P(F) \neq 0$).

If $P(E \cap F) \neq P(E)\,P(F)$, the events are called dependent.

The independence test (the single most-used check in this section):

Step 1 — compute three numbers: $P(E)$, $P(F)$, and $P(E \cap F)$, each directly from the sample space.

Step 2 — compare: multiply $P(E) \cdot P(F)$ and check whether it equals $P(E \cap F)$.

Step 3 — conclude: equal $\Rightarrow$ independent; unequal $\Rightarrow$ dependent.

Worked check. A die is thrown once. Let $E$ = "multiple of 3" $= \{3, 6\}$ and $F$ = "even number" $= \{2, 4, 6\}$.

$$P(E) = \frac{2}{6} = \frac{1}{3}, \qquad P(F) = \frac{3}{6} = \frac{1}{2}, \qquad P(E \cap F) = P(\{6\}) = \frac{1}{6}$$

Since $P(E)\,P(F) = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6} = P(E \cap F)$, the events are independent — even though both are about the same single throw of the same die. Independence is a numerical property, not a statement that the events "come from different experiments".

Independent is NOT the same as mutually exclusive

This is the most-tested conceptual trap of the chapter. The two ideas are not just different — for events with nonzero probability they are incompatible.

If $E$ and $F$ both have nonzero probability and are mutually exclusive, then $P(E \cap F) = 0$ while $P(E)\,P(F) > 0$, so they cannot be independent. Intuitively, mutually exclusive events carry maximum information about each other: if one happened, the other definitely did not.

Board one-liner: "Two events with nonzero probabilities can never be both mutually exclusive and independent." Learn to reproduce the two-line argument above — it appears as a 2-mark reasoning question.

Independent experiments

Often two events sit in different experiments — a coin toss and a die throw, or two separate throws of one die. If the experiments do not influence each other, every event of the first is independent of every event of the second, and we multiply:

$$P(\text{event A in experiment 1 and event B in experiment 2}) = P(A) \cdot P(B)$$

For example, for "a head on the coin and a 3 on the die": $P = \frac{1}{2} \cdot \frac{1}{6} = \frac{1}{12}$. This is why with replacement draws multiply plain probabilities (each draw restarts the experiment identically), while without replacement draws — which change the composition of the box — need the conditional probabilities of the multiplication theorem from the previous section.

Complements, Three Events, and the "At Least One" Formula

Independence survives complements

If $E$ and $F$ are independent, then so are each of the pairs $E$ and $F'$, $\ \ E'$ and $F$, $\ \ E'$ and $F'$. Knowing that $F$ didn't happen is just as uninformative about $E$ as knowing that it did.

Proof for $E$ and $F'$ (a favourite 2–3 mark board proof):

Step 1 — split $E$: the event $E$ is the disjoint union of "$E$ and $F$" and "$E$ and not $F$": $$P(E) = P(E \cap F) + P(E \cap F')$$

Step 2 — solve for the piece we want and use independence: $$P(E \cap F') = P(E) - P(E \cap F) = P(E) - P(E)\,P(F)$$

Step 3 — factor: $$P(E \cap F') = P(E)\,[1 - P(F)] = P(E)\,P(F')$$

which is exactly the independence condition for $E$ and $F'$. $\blacksquare$

Applying this twice gives $P(E' \cap F') = P(E')\,P(F')$ as well.

The "at least one" formula

The complement rule plus independence produces the most useful formula of the section. For independent events $A$ and $B$:

$$P(\text{at least one of } A, B) = P(A \cup B) = 1 - P(A' \cap B') = 1 - P(A')\,P(B')$$

In words: at least one occurs = 1 − (both fail), and "both fail" factors because complements of independent events are independent. The same idea extends to any number of independent events:

$$P(\text{at least one of } A_1, \ldots, A_n) = 1 - P(A_1')\,P(A_2')\cdots P(A_n')$$

Template. "A problem is given to two students who work independently; their chances of solving it are $\frac{1}{2}$ and $\frac{1}{3}$. What is the probability the problem gets solved?"

$$P(\text{solved}) = 1 - P(\text{both fail}) = 1 - \frac{1}{2} \cdot \frac{2}{3} = 1 - \frac{1}{3} = \frac{2}{3}$$

Far faster than adding up the three ways it could be solved.

Three events: mutual independence needs FOUR equations

Three events $A$, $B$, $C$ are mutually independent when all four of these hold:

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

The first three alone are called pairwise independence — and pairwise independence does not guarantee the fourth equation. Conversely, the fourth alone does not guarantee the first three. A JEE-flavoured fact worth remembering: mutual independence is strictly stronger than pairwise independence.

Common mistakes to avoid

Mistake 1 — "independent" read as "mutually exclusive". They are near-opposites for events of positive probability. Mutually exclusive: $P(A \cap B) = 0$. Independent: $P(A \cap B) = P(A)P(B) > 0$.

Mistake 2 — assuming independence without checking. In a single-experiment problem (one die, one card), always run the three-number test. Plausible-sounding pairs frequently fail it: "even number" and "red face" on a coloured die are dependent, even though "even number" and "multiple of 3" are independent.

Mistake 3 — using $P(A \cup B) = P(A) + P(B)$ for independent events. That formula needs mutually exclusive events. For independent events use $P(A \cup B) = P(A) + P(B) - P(A)P(B)$, or the complement route $1 - P(A')P(B')$.

Mistake 4 — claiming three events are mutually independent after checking only pairs. The triple-product equation is a separate requirement.

Solved Examples

Example 1 — The basic independence test

A die is thrown once. $E$ is the event "the number is a multiple of 3" and $F$ is the event "the number is even". Are $E$ and $F$ independent?

Step 1 — list the events: $S = \{1,2,3,4,5,6\}$, $E = \{3, 6\}$, $F = \{2, 4, 6\}$, $E \cap F = \{6\}$.

Step 2 — compute the three numbers: $$P(E) = \frac{2}{6} = \frac{1}{3}, \qquad P(F) = \frac{3}{6} = \frac{1}{2}, \qquad P(E \cap F) = \frac{1}{6}$$

Step 3 — compare: $P(E)\,P(F) = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6} = P(E \cap F)$.

Answer: the product matches, so $E$ and $F$ are independent.

Example 2 — Two throws of a die

An unbiased die is thrown twice. Let $A$ be the event "odd number on the first throw" and $B$ the event "odd number on the second throw". Check whether $A$ and $B$ are independent.

Step 1 — count the sample space: the two throws give $6 \times 6 = 36$ equally likely ordered pairs.

Step 2 — compute each probability: $A$ contains the $3 \times 6 = 18$ pairs with odd first entry, so $P(A) = \frac{18}{36} = \frac{1}{2}$. Similarly $P(B) = \frac{1}{2}$. The event $A \cap B$ (odd on both) contains $3 \times 3 = 9$ pairs, so $P(A \cap B) = \frac{9}{36} = \frac{1}{4}$.

Step 3 — compare: $P(A)\,P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = P(A \cap B)$.

Answer: $A$ and $B$ are independent — as expected, since the two throws do not influence each other.

Example 3 — Three coins, three events: independence must be checked pair by pair

Three coins are tossed simultaneously. Consider the events: $E$ "three heads or three tails", $F$ "at least two heads", $G$ "at most two heads". Of the pairs $(E,F)$, $(E,G)$, $(F,G)$, which are independent and which are dependent?

Step 1 — write the sample space (8 outcomes): $$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$$ $E = \{HHH, TTT\}$, $F = \{HHH, HHT, HTH, THH\}$, $G = S \setminus \{HHH\}$ (everything except three heads).

Step 2 — basic probabilities: $$P(E) = \frac{2}{8} = \frac{1}{4}, \qquad P(F) = \frac{4}{8} = \frac{1}{2}, \qquad P(G) = \frac{7}{8}$$

Step 3 — intersections: $E \cap F = \{HHH\}$ so $P(E \cap F) = \frac{1}{8}$; $E \cap G = \{TTT\}$ so $P(E \cap G) = \frac{1}{8}$; $F \cap G = \{HHT, HTH, THH\}$ (at least two heads but not three) so $P(F \cap G) = \frac{3}{8}$.

Step 4 — test each pair:

$(E, F)$: $\ P(E)P(F) = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8} = P(E \cap F)$ — independent.

$(E, G)$: $\ P(E)P(G) = \frac{1}{4} \cdot \frac{7}{8} = \frac{7}{32} \neq \frac{1}{8}$ — dependent.

$(F, G)$: $\ P(F)P(G) = \frac{1}{2} \cdot \frac{7}{8} = \frac{7}{16} \neq \frac{3}{8}$ — dependent.

Answer: only $E$ and $F$ are independent; the other two pairs are dependent. The moral: independence can hold for some pairs and fail for others within the same experiment, so every pair must be tested on its own.

Example 4 — Independence passes to complements (proof)

If $E$ and $F$ are independent, prove that $E$ and $F'$ are also independent.

Step 1 — decompose $E$: the outcomes in $E$ either lie in $F$ or in $F'$, and these two parts are disjoint: $$P(E) = P(E \cap F) + P(E \cap F')$$

Step 2 — isolate $P(E \cap F')$ and substitute the hypothesis $P(E \cap F) = P(E)P(F)$: $$P(E \cap F') = P(E) - P(E)\,P(F)$$

Step 3 — factor out $P(E)$: $$P(E \cap F') = P(E)\,\big(1 - P(F)\big) = P(E)\,P(F')$$

Answer: $P(E \cap F') = P(E)P(F')$, which is precisely the statement that $E$ and $F'$ are independent. $\blacksquare$

Example 5 — The problem-gets-solved classic

A problem is given to two students $A$ and $B$, who attempt it independently. The probability that $A$ solves it is $\frac{1}{2}$ and the probability that $B$ solves it is $\frac{1}{3}$. Find the probability that (i) the problem is solved, (ii) exactly one of them solves it.

Step 1 — set up complements: $P(A) = \frac{1}{2}$, $P(B) = \frac{1}{3}$, so $P(A') = \frac{1}{2}$, $P(B') = \frac{2}{3}$, and all four pairs formed from $\{A, A'\} \times \{B, B'\}$ are independent.

Step 2 — (i) via "at least one": $$P(\text{solved}) = 1 - P(A')\,P(B') = 1 - \frac{1}{2} \times \frac{2}{3} = 1 - \frac{1}{3} = \frac{2}{3}$$

Step 3 — (ii) exactly one means "$A$ and not $B$" or "$B$ and not $A$" (disjoint cases): $$P(\text{exactly one}) = P(A)\,P(B') + P(B)\,P(A') = \frac{1}{2} \cdot \frac{2}{3} + \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}$$

Answer: (i) $\dfrac{2}{3}$, (ii) $\dfrac{1}{2}$.

Example 6 — Testing independence in a deck of cards

One card is drawn from a well-shuffled deck of 52. In which of the following cases are the events $E$ and $F$ independent? (i) $E$: "spade", $F$: "ace" (ii) $E$: "black", $F$: "king" (iii) $E$: "king or queen", $F$: "king or jack".

Step 1 — case (i): $P(E) = \frac{13}{52} = \frac{1}{4}$, $P(F) = \frac{4}{52} = \frac{1}{13}$, $E \cap F$ = ace of spades, $P(E \cap F) = \frac{1}{52}$. Product: $\frac{1}{4} \cdot \frac{1}{13} = \frac{1}{52}$. ✓ Independent.

Step 2 — case (ii): $P(E) = \frac{26}{52} = \frac{1}{2}$, $P(F) = \frac{4}{52} = \frac{1}{13}$, $E \cap F$ = two black kings, $P(E \cap F) = \frac{2}{52} = \frac{1}{26}$. Product: $\frac{1}{2} \cdot \frac{1}{13} = \frac{1}{26}$. ✓ Independent.

Step 3 — case (iii): $P(E) = \frac{8}{52} = \frac{2}{13}$, $P(F) = \frac{8}{52} = \frac{2}{13}$, $E \cap F$ = the four kings, $P(E \cap F) = \frac{4}{52} = \frac{1}{13}$. Product: $\frac{2}{13} \cdot \frac{2}{13} = \frac{4}{169} \neq \frac{1}{13}$. ✗ Dependent.

Answer: (i) and (ii) independent; (iii) dependent. Note how (iii) fails: the two events share all four kings — half of each event — which is far more overlap than independence would allow.

Example 7 — Percentages as probabilities

In a hostel, 60% of the students read a Hindi newspaper, 40% read an English newspaper and 20% read both. A student is selected at random. (a) Find the probability that she reads neither newspaper. (b) If she reads a Hindi newspaper, find the probability that she reads an English newspaper. (c) If she reads an English newspaper, find the probability that she reads a Hindi newspaper.

Step 1 — translate: $P(H) = 0.6$, $P(E) = 0.4$, $P(H \cap E) = 0.2$.

Step 2 — (a) neither means the complement of "at least one": $$P(H' \cap E') = 1 - P(H \cup E) = 1 - (0.6 + 0.4 - 0.2) = 1 - 0.8 = 0.2 = \frac{1}{5}$$

Step 3 — (b) a conditional, not a product (check: $P(H)P(E) = 0.24 \neq 0.2$, so $H$ and $E$ are not independent and we must condition): $$P(E|H) = \frac{P(E \cap H)}{P(H)} = \frac{0.2}{0.6} = \frac{1}{3}$$

Step 4 — (c) the reverse conditional: $$P(H|E) = \frac{P(H \cap E)}{P(E)} = \frac{0.2}{0.4} = \frac{1}{2}$$

Answer: (a) $\dfrac{1}{5}$, (b) $\dfrac{1}{3}$, (c) $\dfrac{1}{2}$. Parts (b) and (c) have different answers — a live demonstration that $P(E|H) \neq P(H|E)$ in general.