Q.If $P(A) = 0.8$, $P(B) = 0.5$ and $P(B|A) = 0.4$, find

Concept: Conditional Probability — the probability of one event given that another has occurred.

Step 1: Find $P(A \cap B)$

Using the definition of conditional probability:

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

So $P(A \cap B) = P(B|A) \cdot P(A) = 0.4 \times 0.8 = 0.32$.

Step 2: Find $P(A|B)$

$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.32}{0.5} = 0.64$.

Step 3: Find $P(A \cup B)$

Using the addition rule:

$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.8 + 0.5 - 0.32 = 0.98$.

The core idea here is conditional probability — the probability that event $B$ happens given that event $A$ has already occurred. The formula is:

This is not just a formula to plug numbers into; it’s a way of restricting the sample space. When we say “given $A$”, we only care about outcomes where $A$ happens, so the probability of $B$ inside that smaller world is the fraction of $A$ that also contains $B$.

We are given $P(A) = 0.8$, $P(B) = 0.5$, and $P(B|A) = 0.4$. Notice that $P(B|A) = 0.4$ is less than $P(B) = 0.5$, which tells us that $A$ and $B$ are not independent — knowing $A$ actually makes $B$ less likely. That’s a useful sanity check later.

Let’s work through each part step by step.

1. Find $P(A \cap B)$ From the definition of conditional probability:

Multiply both sides by $P(A)$:

So the probability that both $A$ and $B$ occur is $0.32$.

2. Find $P(A|B)$ Now we reverse the conditioning. Using the same definition but swapping roles:

We already have $P(A \cap B) = 0.32$ and $P(B) = 0.5$, so:

Notice that $P(A|B) = 0.64$ is less than $P(A) = 0.8$, consistent with the earlier observation that $A$ and $B$ are negatively associated.

3. Find $P(A \cup B)$ The union probability uses the inclusion-exclusion principle:

Substitute the known values:

This makes sense — since $A$ and $B$ are not mutually exclusive (their intersection is $0.32$, not $0$), the union is less than the sum $1.3$ but still quite high.

Method: Chaining the multiplication, conditional and addition rules

Use this when a question supplies some of $P(A),P(B),P(A\cap B),P(A\mid B),P(B\mid A),P(A\cup B)$ and asks for the rest — you unlock them one formula at a time.

Steps

Step 1: Get the intersection first (multiplication rule)

Almost every other quantity needs $P(A\cap B)$. Rearrange a given conditional probability into a product:

Use whichever conditional the question actually gives.

Step 2: Compute any conditional you still need

Match the denominator to the event after the bar — swapping them is the classic error.

Step 3: Combine into the union (addition rule)

Subtracting the intersection removes the double-count of the overlap.

Step 4: Cross-check independence

Independent events would satisfy $P(A\cap B)=P(A)P(B)$; if the computed intersection differs, the events are simply dependent — a useful consistency read, not a mistake.

Common Mistakes

Mistake 1: Assuming independence, $P(A\cap B)=P(A)\,P(B)=0.8\times0.5=0.40$.

Why it's wrong: $P(B\mid A)=0.4\neq P(B)=0.5$, so $A$ and $B$ are not independent; you must use $P(A\cap B)=P(B\mid A)\,P(A)=0.4\times0.8=0.32$. The independence product $0.40$ is simply wrong here.

Mistake 2: Recovering $P(A\cap B)$ as $P(B\mid A)\,P(B)$ instead of $P(B\mid A)\,P(A)$.

Why it's wrong: since $P(B\mid A)=\dfrac{P(A\cap B)}{P(A)}$, the multiplication rule multiplies by $P(A)$ — the event being conditioned on. Correct approach: $0.4\times0.8=0.32$.

Mistake 3: Confusing $P(A\mid B)$ with the given $P(B\mid A)$.

Why it's wrong: they have different denominators; $P(A\mid B)=\dfrac{0.32}{0.5}=0.64$ while $P(B\mid A)=0.4$. Reusing $0.4$ for part (ii) is wrong.