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

Options

A. `1/5`
B. 0
C. `1/3`
D. `1/2`

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the probabilities
Let:
- \( P(H) \) = Probability of a student reading Hindi newspapers = 60% = 0.6
- \( P(E) \) = Probability of a student reading English newspapers = 40% = 0.4
- \( P(H \cap E) \) = Probability of a student reading both Hindi and English newspapers = 20% = 0.2

### Step 2: Use the formula for union of probabilities
We need to find the probability that a student reads neither Hindi nor English newspapers, which can be expressed as:
\[ P(H' \cap E') = 1 - P(H \cup E) \]
Where \( P(H \cup E) \) is the probability that a student reads either Hindi or English newspapers.

Using the formula for the union of two probabilities:
\[ P(H \cup E) = P(H) + P(E) - P(H \cap E) \]

### Step 3: Substitute the values into the formula
Now, substituting the known values:
\[ P(H \cup E) = P(H) + P(E) - P(H \cap E) \]
\[ P(H \cup E) = 0.6 + 0.4 - 0.2 \]
\[ P(H \cup E) = 1.0 - 0.2 = 0.8 \]

### Step 4: Calculate the probability of neither reading
Now, we can find the probability that a student reads neither Hindi nor English newspapers:
\[ P(H' \cap E') = 1 - P(H \cup E) = 1 - 0.8 = 0.2 \]

### Step 5: Convert to fraction
The probability \( 0.2 \) can be expressed as a fraction:
\[ P(H' \cap E') = \frac{2}{10} = \frac{1}{5} \]

### Final Answer
Thus, the probability that a randomly selected student reads neither Hindi nor English newspapers is:
\[ \frac{1}{5} \]