Q.If $P(A) = \frac{6}{11}$, $P(B) = \frac{5}{11}$ and $P(A \cup B) = \frac{7}{11}$, find

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

Step 1 – Find $P(A \cap B)$

Using the addition theorem:

Substitute:

So:

Step 2 – Find $P(A|B)$

$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{4/11}{5/11} = \frac{4}{5}$ …

The core idea here is conditional probability — the chance of one event happening given that another has already occurred. The formula is simple:

But to use it, we first need $P(A \cap B)$, the probability that both A and B occur. That’s where the addition rule comes in: it connects the union, the individual probabilities, and the intersection.

Let’s walk through it.

1. Find $P(A \cap B)$ using the addition rule

   The addition rule for any two events is:

Plug in the given values:

Simplify the right side:

So:

Rearranging:

2. Find $P(A|B)$

   Using the definition:

   $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{4/11}{5/11} = \frac{4}{5}$ …

Method: From the union back to the intersection, then both conditionals

Use this when $P(A)$, $P(B)$ and the union $P(A\cup B)$ are known and you must recover the intersection and the conditional probabilities.

Steps

Step 1: Rearrange the addition rule for the intersection

The addition rule solved for the overlap gives

This is the key quantity every conditional below depends on.

Step 2: Form each conditional by dividing

$P(A\mid B)=\frac{P(A\cap B)}{P(B)},\qquad P(B\mid A)=\frac{P(A\cap B)}{P(A)}.$ …

Common Mistakes

Mistake 1: Swapping $P(A\mid B)$ and $P(B\mid A)$ — using the same denominator for both.

Why it's wrong: they divide by different events: $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{4/11}{5/11}=\dfrac45$, but $P(B\mid A)=\dfrac{P(A\cap B)}{P(A)}=\dfrac{4/11}{6/11}=\dfrac23$. They are equal only if $P(A)=P(B)$.

Mistake 2: Sign error when extracting $P(A\cap B)$ from the addition rule. …