Independent events

Two events $A$ and $B$ are independent if knowing whether one occurs gives no information about the other. Formally: $P(A \cap B) = P(A) \cdot P(B).$ Equivalently (when $P(B) > 0$): $P(A \mid B) = P(A)$. The conditional probability equals the unconditional probability , conditioning on $B$ doesn't change $A$'s probability.

Three equivalent definitions

The following are equivalent for events with positive probabilities:

1. $P(A \cap B) = P(A) P(B)$.
2. $P(A \mid B) = P(A)$.
3. $P(B \mid A) = P(B)$.

Use whichever is most convenient for the situation. The first is the official definition because it doesn't require any probability to be positive.

Independence vs disjointness

These are different things, and easily confused. Disjoint events have empty intersection: $A \cap B = \emptyset$. Independent events have $P(A \cap B) = P(A) P(B)$.

Two disjoint events with positive probabilities are not independent (because if one occurs, the other cannot , so $P(B \mid A) = 0 \ne P(B)$). The only way to be both disjoint and independent is if at least one has probability zero.

Pairwise vs mutual independence

For three events $A, B, C$ to be pairwise independent: each pair is independent. For mutual independence: every subset is independent, i.e. $P(A \cap B) = P(A)P(B)$, $P(B \cap C) = P(B)P(C)$, $P(A \cap C) = P(A)P(C)$, and $P(A \cap B \cap C) = P(A)P(B)P(C)$.

Pairwise independence does not imply mutual independence. A classic example: roll two dice. $A =$ first is odd, $B =$ second is odd, $C =$ sum is odd. Pairwise independent , but $A, B$ together determine $C$, so the three are not mutually independent.

In Class XII, "independent" usually means mutually independent for the events in question.

Independence of complements

If $A$ and $B$ are independent, so are $A^c$ and $B$, $A$ and $B^c$, $A^c$ and $B^c$. The argument: $P(A^c \cap B) = P(B) - P(A \cap B) = P(B)(1 - P(A)) = P(A^c)P(B)$.

Independent trials

When experiments are repeated independently (coin flips, die rolls, sampling with replacement), the events from each trial are independent. So P(\text{event in all n trials}) = P(\text{event in 1 trial})^n.

Worked examples

Example 1. Flip a fair coin twice. $A =$ first is heads, $B =$ second is heads. Are they independent?

$P(A) = 1/2 = P(B)$. $P(A \cap B) = P(\text{HH}) = 1/4 = P(A)P(B)$. Yes, independent.

Example 2. Roll a die. $A =$ even, $B =$ at least 4. Are they independent?

$P(A) = 1/2$, $P(B) = 1/2$. $A \cap B = \{4, 6\}$, $P(A \cap B) = 2/6 = 1/3$. But $P(A)P(B) = 1/4 \ne 1/3$. Not independent.

Example 3. Two coins are flipped. Find the probability of exactly one head.

$P(\text{HT}) + P(\text{TH}) = (1/2)(1/2) + (1/2)(1/2) = 1/2$. (Used independence of the two flips.)

Example 4. $P(A) = 0.4, P(B) = 0.5$, and $A, B$ are independent. Find $P(A \cup B)$.

$P(A \cap B) = 0.2$. $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.5 - 0.2 = 0.7$.

Example 5. Three independent shooters each hit a target with probability $0.6, 0.7, 0.5$. Find the probability that at least one hits.

$P(\text{none hits}) = 0.4 \cdot 0.3 \cdot 0.5 = 0.06$. So $P(\text{at least one hits}) = 1 - 0.06 = 0.94$.

Example 6. A fair die is rolled three times. Find the probability that all three rolls are different.

$P = 1 \cdot (5/6) \cdot (4/6) = 20/36 = 5/9$. The roll outcomes are independent; the constraint "different" multiplies through.

Try it yourself

1. Flip a coin three times. Find $P(\text{all heads})$.
2. $P(A) = 0.3, P(B) = 0.6$, independent. Find $P(A \cap B)$ and $P(A \cup B)$.
3. Two fair dice rolled. Are $A =$ first is even and $B =$ second is even independent?
4. A coin is flipped 4 times. Find $P(\text{first 2 are heads, last 2 are tails})$.
5. $P(A) = 0.5, P(B) = 0.4$, $P(A \cup B) = 0.7$. Are $A, B$ independent?
6. Independent events with $P(A) = 0.6, P(B) = 0.4$. Find $P(A^c \cap B)$.
7. Three students independently solve a problem with probabilities $0.7, 0.5, 0.8$. Find $P(\text{all solve})$.
8. Same as previous: find $P(\text{exactly one solves})$.
9. Two cards from a deck with replacement: $P(\text{both aces})$.
10. Same but without replacement: compute and compare with previous. (Hint: the second case is not independent.)
11. A box has 3 white, 7 black balls. Draw two with replacement: $P(\text{both white})$.
12. Same box, without replacement: $P(\text{both white})$.
13. Independent $A, B$ with $P(A) = 0.4$, $P(B) = 0.5$. Find $P(\text{neither occurs})$.
14. Roll a die three times. Find $P(\text{at least one 6})$.

Pitfalls and tricks

• Test for independence with $P(A \cap B) = P(A) P(B)$. Don't conflate independent with disjoint.
• With-replacement sampling is independent; without-replacement is not.
• Probability of at least one occurrence: compute the complement (none occurs) and subtract from 1.
• For multiple independent events, the probability of all occurring is the product.
• Complement rule applies within independence: $P(A^c) = 1 - P(A)$, but you also have $A^c$ independent of $B$ if $A$ is.