In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random. (a) Find the probability that she reads neither Hindi nor English newspapers. (b) If she reads Hindi newspaper, find

Question

In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random. (a) Find the probability that she reads neither Hindi nor English newspapers. (b) If she reads Hindi newspaper, find the probability that she reads English newspaper. (c) If she reads English newspaper, find the probability that she reads Hindi newspaper.

Accepted Answer

> [!TLDR]
> This problem uses the Inclusion-Exclusion Principle to handle overlapping events. Given $P(H)=0.6$, $P(E)=0.4$, $P(H\cap E)=0.2$, we find (a) $P(	ext{neither})=0.2$,
> (b) $P(E|H)=\frac{1}{3}$,
> (c) $P(H|E)=\frac{1}{2}$.

The core idea here is that reading Hindi and reading English are **not** mutually exclusive — 20% of students read both. When events overlap, you cannot simply add or subtract probabilities without accounting for the double-counted intersection. That’s exactly where the Inclusion-Exclusion Principle steps in.

For any two events $A$ and $B$:
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
This formula ensures we count the overlap only once. Once we know $P(A \cup B)$, the probability of “neither” is simply $1 - P(A \cup B)$.

For parts (b) and (c), we need **conditional probability** — the chance of one event given that the other has already happened. The definition is:
$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$
This shrinks the “universe” to only those who satisfy the condition.

Let’s apply these ideas step by step.

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1. **Define the events clearly**  
   Let $H$ = “student reads Hindi newspaper” and $E$ = “student reads English newspaper”.  
   From the problem:  
   $$P(H) = 0.6, \quad P(E) = 0.4, \quad P(H \cap E) = 0.2$$

2. **Find the probability that a student reads at least one newspaper**  
   Using Inclusion-Exclusion:  
   $$P(H \cup E) = P(H) + P(E) - P(H \cap E) = 0.6 + 0.4 - 0.2 = 0.8$$  
   So 80% of students read Hindi or English (or both).

3. **Part (a): Probability of reading neither**  
   “Neither” is the complement of “at least one”:  
   $$P(	ext{neither}) = 1 - P(H \cup E) = 1 - 0.8 = 0.2$$  
   > [!WARNING]
   > A common mistake is to subtract $P(H \cap E)$ twice — e.g., doing $1 - P(H) - P(E) + P(H \cap E)$ is actually correct but only if you remember the plus sign. The cleanest path is to find $P(H \cup E)$ first, then complement. …

Q.In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random.

Concept understanding — Conditional Probability

Conditional Probability

The Idea: Shrink the Sample Space

The Definition

Two Cautions

Method: Overlapping-group problems — addition rule, complement, and conditioning together

Steps

Common Mistakes

Step 1 — identify the probabilities

Step 2 — apply the conditional-probability formula

Step 1 — list the outcomes with sum 7.

Step 2 — count first die greater than second, among those.

Step 3 — conditional probability.

Step 1 — Find the joint probability

Step 2 — Apply the definition of conditional probability

Step 1 — Intersection via the addition rule

Step 2 — Apply the conditional-probability formula

Step 1 — Apply the conditional probability formula for $P(A\mid B)$

Step 2 — Apply it for $P(B\mid A)$

Step 3 — Sanity check

Step 1 — Identify the reduced sample space

Step 2 — Apply the conditional formula

Q.In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random.

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