Q.In a college, $30\%$ students fail in physics, $25\%$ fail in mathematics and $10\%$ fail in both. One student is chosen at random. The probability that she fails in physics if she has failed in mathematics is
(A) $\dfrac{1}{10}$
(B) $\dfrac{2}{5}$
(C) $\dfrac{9}{20}$
(D) $\dfrac{1}{3}$

Concept: Conditional Probability — the probability of event $A$ given $B$ is $P(A|B) = \frac{P(A \cap B)}{P(B)}$.

Let $P$ = fails in Physics, $M$ = fails in Mathematics.

Given: $P(P) = 0.30$, $P(M) = 0.25$, $P(P \cap M) = 0.10$.

We need $P(P|M)$.

Step 1: Write the formula:

$P(P|M) = \frac{P(P \cap M)}{P(M)}$ …

The question asks: given that a student has already failed mathematics, what is the chance she also fails physics? This is a textbook conditional probability problem. The key is to realise that the condition ("if she has failed in mathematics") shrinks our sample space — we are no longer considering all students, only those who failed maths. Within that smaller group, we want the fraction who also failed physics.

Conditional probability is defined as:

where $A$ is the event "fails physics" and $B$ is the event "fails mathematics". The numerator is the probability of both events happening together; the denominator is the probability of the condition.

Let's work through the numbers.

1. Identify the given probabilities from the problem.

   • $30\%$ fail physics: $P(\text{Physics fail}) = 0.30$
   • $25\%$ fail mathematics: $P(\text{Maths fail}) = 0.25$
   • $10\%$ fail both: $P(\text{Both fail}) = 0.10$

2. Write down what we need.

   We want $P(\text{Physics fail} \mid \text{Maths fail})$. Using the formula:

3. Plug in the values. The numerator is $0.10$ (both fail), and the denominator is $0.25$ (fails maths). So:

4. Interpret the result. …

Method: Conditional probability from percentage / overlap data

Use this when you are given $P(A)$, $P(B)$ and the overlap $P(A\cap B)$ (often as percentages) and asked for $P(A \mid B)$.

Steps

Step 1: Name the events and read off the three probabilities.

Identify $P(A)$, $P(B)$ and especially $P(A\cap B)$ — the "both" figure — converting percentages to fractions or decimals.

Step 2: Write the conditional-probability formula.

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

Common Mistakes

Mistake 1: Dividing by $P(\text{physics})=0.30$ instead of $P(\text{maths})=0.25$.

Why it's wrong: the condition is "has failed mathematics", so the denominator must be $P(\text{maths})$. Correct approach: $P(\text{physics}\mid\text{maths})=\dfrac{P(\text{both})}{P(\text{maths})}=\dfrac{0.10}{0.25}=\dfrac{2}{5}$.

Mistake 2: Answering with $P(\text{physics})=0.30$ directly. …