A number is chosen at random from first ten natural numbers. The probability that number is odd and perfect square is .

Options

A. `(2)/(9)`
B. `(2)/(5)`
C. `(3)/(7)`
D. `(1)/(5)`

Answer

To solve the problem, we need to find the probability that a randomly chosen number from the first ten natural numbers is both odd and a perfect square.

### Step-by-Step Solution:

1. **Identify the Total Cases**:
The first ten natural numbers are:
\[ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \]
The total number of cases (total natural numbers) is 10.

2. **Identify the Favorable Cases**:
We need to find numbers that are both odd and perfect squares.
- First, list the odd numbers from 1 to 10:
\[ 1, 3, 5, 7, 9 \]
- Now, identify which of these odd numbers are perfect squares:
- \(1 = 1^2\) (perfect square)
- \(3\) (not a perfect square)
- \(5\) (not a perfect square)
- \(7\) (not a perfect square)
- \(9 = 3^2\) (perfect square)

The odd perfect squares from the list are:
\[ 1 \text{ and } 9 \]
Therefore, the favorable cases are \(1\) and \(9\), giving us a total of 2 favorable cases.

3. **Calculate the Probability**:
The probability \(P\) is given by the formula:
\[ P = \frac{\text{Number of Favorable Cases}}{\text{Total Cases}} = \frac{2}{10} = \frac{1}{5} \]

### Final Answer:
The probability that the number chosen is odd and a perfect square is:
\[ \frac{1}{5} \]