Q.Given that the events A and B are such that $P(A) = \frac{1}{2}$, $P(A \cup B) = \frac{3}{5}$ and $P(B) = p$. Find $p$ if they are

Concept: Event Independence — For independent events, $P(A \cap B) = P(A)P(B)$. For mutually exclusive events, $P(A \cap B) = 0$.

1. Mutually exclusive: $P(A \cup B) = P(A) + P(B)$ because $P(A \cap B) = 0$. So $\frac{3}{5} = \frac{1}{2} + p \implies p = \frac{3}{5} - \frac{1}{2} = \frac{6-5}{10} = \frac{1}{10}$.
2. Independent: $P(A \cup B) = P(A) + P(B) - P(A)P(B)$. …

The key to this problem is understanding the difference between two fundamental relationships between events: mutual exclusivity and independence. These are often confused by students, but they describe completely different things.

Mutually exclusive means the events cannot happen at the same time — their intersection is empty. Independence means the occurrence of one does not affect the probability of the other — their intersection probability is the product of their individual probabilities.

Let's work through each case.

1. Mutually exclusive case

When A and B are mutually exclusive, $A \cap B = \emptyset$, so $P(A \cap B) = 0$.

The addition rule for any two events is:

Since $P(A \cap B) = 0$, this simplifies to:

Substitute the given values:

Solve for $p$:

2. Independent case

When A and B are independent, the probability of their intersection is the product of their individual probabilities:

Now use the full addition rule:

Substitute: …

Method: Solving for an Unknown Probability with the Addition Rule

Use this when you are given $P(A)$, $P(A\cup B)$ and a relationship between the events (mutually exclusive or independent), and must find the missing $P(B)$.

Steps

Step 1: Start from the general addition rule

For any two events,

Step 2: Replace $P(A\cap B)$ using the stated relationship

• Mutually exclusive: the events cannot co-occur, so $P(A\cap B)=0$.
• Independent: use the product rule, $P(A\cap B)=P(A)\,P(B)$.

Substitute the appropriate one — this is the step that distinguishes the two parts. …

Common Mistakes

Mistake 1: Swapping the two intersection rules.

Students use $P(A\cap B)=P(A)P(B)$ for the mutually-exclusive part or $P(A\cap B)=0$ for the independent part. Why it's wrong: mutually exclusive means $P(A\cap B)=0$; independent means $P(A\cap B)=P(A)P(B)$ — opposite situations. Correct approach: mutually exclusive $\Rightarrow\frac{3}{5}=\frac12+p\Rightarrow p=\frac{1}{10}$; independent $\Rightarrow\frac{3}{5}=\frac12+p-\frac12 p\Rightarrow p=\frac{1}{5}$. …