Q.Given two independent events A and B such that $P(A) = 0.3$, $P(B) = 0.6$. Find

Concept: Event Independence — For independent events, the probability of both occurring is the product of their individual probabilities.

Step 1: $P(A \text{ and } B) = P(A) \cdot P(B) = 0.3 \times 0.6 = 0.18$.

Step 2: $P(A \text{ and not } B) = P(A) \cdot P(\text{not } B) = 0.3 \times (1 - 0.6) = 0.3 \times 0.4 = 0.12$.

Step 3: $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) = 0.3 + 0.6 - 0.18 = 0.72$. …

The key idea here is event independence. When two events are independent, the occurrence of one does not affect the probability of the other. This gives us a clean, multiplicative rule for their intersection — no conditional probabilities needed.

For independent events $A$ and $B$:

• $P(A \cap B) = P(A) \cdot P(B)$
• $P(A \cap B') = P(A) \cdot P(B')$ (since $A$ and $B'$ are also independent)
• $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ (the general addition rule)
• $P(\text{neither } A \text{ nor } B) = P(A' \cap B') = 1 - P(A \cup B)$

Let's work through each part step by step.

1. Find $P(A \text{ and } B)$, i.e., $P(A \cap B)$ Since $A$ and $B$ are independent:

2. Find $P(A \text{ and not } B)$, i.e., $P(A \cap B')$ Here $B'$ is the complement of $B$, with $P(B') = 1 - 0.6 = 0.4$. Independence of $A$ and $B$ implies $A$ is also independent of $B'$ (this is a standard result — if two events are independent, each is independent of the other's complement). So:

3. Find $P(A \text{ or } B)$, i.e., $P(A \cup B)$ Use the addition rule for any two events:

Substituting the values:

$P(A \cup B) = 0.3 + 0.6 - 0.18 = 0.72$ …

Method: Building compound probabilities from two independent events

Once two events are known to be independent, every "and / and-not / or / neither" combination follows from one rule plus complements — no conditional probability needed.

Steps

Step 1: For an intersection, multiply.

Independence also carries to complements, so $P(A \cap B') = P(A)\,P(B')$ with $P(B') = 1 - P(B)$.

Step 2: For a union, use the addition rule.

using the product from Step 1 for the overlap. …

Common Mistakes

Mistake 1: For "$A$ and not $B$", subtracting probabilities instead of using independence with the complement.

Why it's wrong: $A$ is independent of $B'$ too, so $P(A\cap B')=P(A)P(B')=0.3\times0.4=0.12$ — not $P(A)-P(B)$. Correct approach: multiply by $P(B')=1-P(B)$. …