Q.Three events $A$, $B$ and $C$ have probabilities $\dfrac{2}{5}$, $\dfrac{1}{3}$ and $\dfrac{1}{2}$, respectively. Given that $P(A \cap C) = \dfrac{1}{5}$ and $P(B \cap C) = \dfrac{1}{4}$, find the values of $P(C \mid B)$ and $P(A' \cap C')$.

Concept: Conditional Probability and Complement Rule

Step 1 – Find $P(C \mid B)$

By definition:

Step 2 – Find $P(A' \cap C')$

Using De Morgan’s law: $A' \cap C' = (A \cup C)'$, so

Step 3 – Compute $P(A \cup C)$

$P(A \cup C) = P(A) + P(C) - P(A \cap C) = \frac{2}{5} + \frac{1}{2} - \frac{1}{5}$ …

We start with the definition of conditional probability. For any two events $X$ and $Y$, the probability of $X$ given $Y$ is

provided $P(Y) > 0$. This is not a formula to memorise blindly — it makes sense: if we know $Y$ has happened, we restrict our attention to that part of the sample space, and we want the fraction of $Y$ that also contains $X$.

For the second part, $P(A' \cap C')$ is the probability that neither $A$ nor $C$ occurs. By De Morgan’s law, $A' \cap C' = (A \cup C)'$, so

We already have $P(A)$, $P(C)$, and $P(A \cap C)$, so we can find $P(A \cup C)$ using the addition rule.

Let’s work through it step by step.

1. Find $P(C \mid B)$ Using the definition:

We are given $P(B \cap C) = \frac{1}{4}$ and $P(B) = \frac{1}{3}$.

So

2. Find $P(A' \cap C')$ First, compute $P(A \cup C)$ using the addition rule:

Substitute the given values:

Get a common denominator (10): …

Method: Conditional Probability and the "Neither" Event via De Morgan

Use this when a problem mixes a conditional probability with a "neither event occurs" probability.

Steps

Step 1: Compute the conditional directly.

Only the joint $P(B\cap C)$ and the conditioning $P(B)$ are needed — ignore data the question doesn't require.

Step 2: Translate "neither" with De Morgan. …

Common Mistakes

Mistake 1: Computing $P(A'\cap C')$ as $1-P(A)-P(C)$.

Why it's wrong: that ignores the overlap $A\cap C$; "neither" is $1-P(A\cup C)$. Correct approach: $P(A\cup C)=P(A)+P(C)-P(A\cap C)=\frac{7}{10}$, so $P(A'\cap C')=\frac{3}{10}$.

Mistake 2: Assuming $A$ and $C$ are independent to get $P(A\cap C)$. …