Understanding Probability of Independent Events

The probability of independent events addresses scenarios where the occurrence of one event does not affect the probability of another. The analysis involves understanding definitions, notation, key properties, calculation principles, and illustrative problems relevant for JEE Main examinations.

Definition and Notation for Independence of Events

Definition: Events $A$ and $B$ in a sample space are independent if the occurrence of $A$ does not alter the probability of $B$, and vice versa. Algebraically, $P(A \cap B) = P(A) \cdot P(B)$ represents the independence of $A$ and $B$.

For more than two events, $A_1, A_2, ..., A_n$ are mutually independent if, for every subset $\{i_1, ..., i_k\}$, $P(A_{i_1} \cap \cdots \cap A_{i_k}) = P(A_{i_1}) \cdots P(A_{i_k})$ for $2 \leq k \leq n$.

The notation $P(A \mid B)$ denotes the conditional probability of $A$ given $B$. For independent events, $P(A \mid B) = P(A)$ provided $P(B) > 0$.

Multiplication Principle in Calculating Joint Probability

Result: For independent events $A$ and $B$, the probability of their simultaneous occurrence equals the product of their individual probabilities: $P(A \cap B) = P(A) \cdot P(B)$. This extends to $n$ independent events as $P(A_1 \cap \ldots \cap A_n) = \prod_{i=1}^n P(A_i)$.

This property allows rapid evaluation of probabilities involving multiple independent steps, such as sequential tosses of coins or rolls of dice. Refer to the Multiplication Theorem Of Probability for the formal theorem.

Differentiating Independent Events and Mutually Exclusive Events

Common Error: Students frequently confuse independence with mutual exclusivity. Independent events can occur simultaneously, whereas mutually exclusive events cannot.

Mutually exclusive events satisfy $P(A \cap B) = 0$, and so are dependent except in the trivial case. For distinction, consult the Difference Between Independent And Dependent Events resource.

Conditional Probability in the Context of Independence

For events $A$ and $B$, the conditional probability formula $P(A | B) = \dfrac{P(A \cap B)}{P(B)}$ is structurally simplified for independent events. If $P(B) > 0$, then $P(A | B) = P(A)$, reflecting that the knowledge of $B$ does not update the probability of $A$.

Similarly, if $P(A) > 0$, $P(B | A) = P(B)$. This property validates the independence assumption in problem statements where trials or processes are physically distinct.

Typical JEE Patterns Involving Probability of Independent Events

• Tossing unbiased coins multiple times
• Rolling dice together
• Simultaneous but distinct experiments
• Sequenced independent selection events
• Joint occurrence of independent outcomes

Frequently, questions require calculation of the probability that a combination of independent events all occur or none occur. See Experimental Vs Theoretical Probability for linkages to types of probability.

Illustrative Problems on Probability of Independent Events

Example: Three unbiased coins are tossed. Find the probability that the first shows head, the second shows tail, and the third shows head.

Solution: Probability (first coin is head) $= \dfrac{1}{2}$. Probability (second coin is tail) $= \dfrac{1}{2}$. Probability (third coin is head) $= \dfrac{1}{2}$. By independence, the required probability is $P = \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{8}$.

Example: A die is rolled and a coin is tossed. Find the probability that the die shows a $2$ and the coin shows tail.

Solution: Probability (die shows $2$) $= \dfrac{1}{6}$. Probability (coin shows tail) $= \dfrac{1}{2}$. Since the experiments are independent, $P = \dfrac{1}{6} \times \dfrac{1}{2} = \dfrac{1}{12}$.

Example: The chance that a flight is not delayed is $0.8$. What is the probability that both onward and return flights are not delayed?

Solution: Required probability is $0.8 \times 0.8 = 0.64$.

For more advanced examples, including intersection and union of independent events, see Understanding Probability.

Common Misconceptions: The Gambler’s Fallacy and Related Errors

Exam Tip: A prevalent error is to assume past outcomes influence future ones in independent trials (for example, expecting a tail after successive heads in coin tosses). Probabilities remain unchanged; every toss is independent. This error is called the Gambler’s Fallacy.

Students must evaluate each independent repetition or parallel experiment distinctly, without adjustment for previous outcomes. See Dependent Events In Probability for cases where such adjustment is required.