Q.Fill in the blank: If $A$ and $B$ are two events such that $P(A \mid B) = p$, $P(A) = p$, $P(B) = \dfrac{1}{3}$ and $P(A \cup B) = \dfrac{5}{9}$, then $p =$ __________.

Concept: Probability Complement Rule — Use $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ and the given conditional probability to find $p$.

Step 1: From $P(A \mid B) = p$, we have

$P(A \cap B) = P(A \mid B) \cdot P(B) = p \cdot \frac{1}{3}$.

Step 2: Substitute into the union formula:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$\frac{5}{9} = p + \frac{1}{3} - \frac{p}{3}$.

Step 3: Solve: …

Let’s understand why this works before jumping into algebra. The problem gives us $P(A \mid B) = p$ and $P(A) = p$. When the conditional probability of $A$ given $B$ equals the unconditional probability of $A$, it tells us something special: knowing whether $B$ happened gives no information about $A$. That is the definition of independence between $A$ and $B$.

So the moment we see $P(A \mid B) = P(A)$, we know independence holds. This is the central insight that unlocks the problem.

Now let’s work through it step by step.

1. Write the independence condition. Since $P(A \mid B) = p$ and $P(A) = p$, we have $P(A \mid B) = P(A)$. Therefore $A$ and $B$ are independent. For independent events, the probability of their intersection is the product of their probabilities:

2. Use the union probability formula. We are given $P(A \cup B) = \frac{5}{9}$. The general addition rule says:

Substitute what we know:

3. Solve for $p$. First, simplify the right-hand side. Write $\frac{1}{3}$ as $\frac{3}{9}$ to keep denominators consistent later, but let’s solve algebraically first:

Combine the $p$ terms: $p - \frac{p}{3} = \frac{2p}{3}$. So:

Subtract $\frac{1}{3}$ from both sides. Note $\frac{1}{3} = \frac{3}{9}$: …

Method: Using the independence clue $P(A\mid B)=P(A)$ with the addition rule

Use this when a problem hands you $P(A\mid B)=P(A)$ (possibly disguised, e.g. both equal $p$) together with union/intersection data.

Steps

Step 1: Read $P(A\mid B)=P(A)$ as independence.

When conditioning on $B$ leaves $P(A)$ unchanged, $A$ and $B$ are independent, so

Step 2: Write the addition (union) rule.

Step 3: Substitute the independent intersection and the known values. …

Common Mistakes

Mistake 1: Missing that $P(A\mid B)=P(A)$ signals independence.

Why it's wrong: not spotting this leaves $P(A\cap B)$ unknown and the equation unsolvable in one variable. Correct approach: independence gives $P(A\cap B)=P(A)P(B)=\tfrac{p}{3}$.

Mistake 2: Writing $P(A\cap B)=P(A\mid B)\cdot P(A)$ instead of $P(A\mid B)\cdot P(B)$. …