Q.$A$ and $B$ are two events such that $P(A) = \dfrac{1}{2}$, $P(B) = \dfrac{1}{3}$ and $P(A \cap B) = \dfrac{1}{4}$. Find:

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

Step 1 – Formula

$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$, provided $P(B) \neq 0$.

Step 2 – Compute (i) and (ii)

1. $P(A \mid B) = \frac{1/4}{1/3} = \frac{3}{4}$
2. $P(B \mid A) = \frac{1/4}{1/2} = \frac{1}{2}$ Step 3 – Compute (iii) $P(A' \mid B) = 1 - P(A \mid B) = 1 - \frac{3}{4} = \frac{1}{4}$ Step 4 – Compute (iv) First, $P(B') = 1 - \frac{1}{3} = \frac{2}{3}$, and $P(A' \cap B') = 1 - P(A \cup B)$. …

Why conditional probability works here

Conditional probability answers the question: If we know that event $B$ has happened, how does that change the chance of event $A$? The key idea is that knowing $B$ occurred shrinks the "universe" of possible outcomes from the whole sample space down to just $B$. So the probability of $A$ given $B$ is the proportion of $B$ that also belongs to $A$ — that is, $P(A \cap B)$ divided by $P(B)$.

The formula is symmetric in logic: $P(A|B) = \frac{P(A \cap B)}{P(B)}$, provided $P(B) \neq 0$. For the complement events, we use the fact that $P(A' \cap B) = P(B) - P(A \cap B)$, since $A' \cap B$ is just the part of $B$ that lies outside $A$.

Let's work through each part.

1. Finding $P(A \mid B)$

We have $P(A \cap B) = \frac{1}{4}$ and $P(B) = \frac{1}{3}$.

So given that $B$ occurred, there is a $\frac{3}{4}$ chance that $A$ also occurred.

2. Finding $P(B \mid A)$

Now we condition on $A$ instead. $P(A) = \frac{1}{2}$.

Notice that $P(B|A) = \frac{1}{2}$ is larger than $P(B) = \frac{1}{3}$, which tells us that $A$ and $B$ are positively associated — knowing $A$ happened makes $B$ more likely.

3. Finding $P(A' \mid B)$

Here we want the probability that $A$ does not occur, given that $B$ has occurred. The event $A' \cap B$ is the part of $B$ that is outside $A$.

Now divide by $P(B)$:

4. Finding $P(A' \mid B')$

Now we condition on $B$ not happening. First, find $P(B')$: …

Method: Computing a Family of Conditional Probabilities Efficiently

Use this when one setup asks for several conditionals such as $P(A\mid B)$, $P(B\mid A)$, $P(A'\mid B)$, $P(A'\mid B')$.

Steps

Step 1: Start from the definition with the conditioning event below.

Step 2: For a complement of the numerator under the same condition, use the shortcut.

Since $A$ and $A'$ partition, conditional probabilities on the same event add to $1$:

$P(A'\mid B)=1-P(A\mid B).$ …

Common Mistakes

Mistake 1: Writing $P(A'|B')=1-P(A|B)$.

Why it's wrong: the $1-(\cdot)$ shortcut only works when the conditioning event is unchanged; here the condition switches from $B$ to $B'$. Correct approach: $P(A'|B')=\frac{P(A'\cap B')}{P(B')}=\frac{5/12}{2/3}=\frac58$.

Mistake 2: Getting $P(A'\cap B')$ from $1-P(A)-P(B)$. …