Number 1, 2, 3,…,2n (n in N) are printed on 2n cards. The probability of drawing a number r is proportional to r. Then the probability of drawing an even number in one draw is

Options

A. `(n+2)/(n+3)`
B. `(n+1)/(n+3)`
C. `(1)/(2)`
D. `(n+1)/(2n+1)`

Answer

To solve the problem, we need to find the probability of drawing an even number from the cards numbered 1 to 2n, where the probability of drawing a number \( r \) is proportional to \( r \).

### Step-by-Step Solution:

1. **Understanding the Probability Distribution**:
The probability of drawing a number \( r \) is given as proportional to \( r \). Therefore, we can express this as:
\[ P(X = r) = k \cdot r \]
where \( k \) is a constant of proportionality.

2. **Total Probability Must Equal 1**:
The sum of probabilities for all possible outcomes must equal 1:
\[ \sum_{r=1}^{2n} P(X = r) = 1 \]
Substituting the expression for \( P(X = r) \):
\[ \sum_{r=1}^{2n} k \cdot r = 1 \]
This can be rewritten as:
\[ k \sum_{r=1}^{2n} r = 1 \]

3. **Calculating the Sum of the First \( 2n \) Natural Numbers**:
The sum of the first \( m \) natural numbers is given by the formula:
\[ \sum_{r=1}^{m} r = \frac{m(m + 1)}{2} \]
For \( m = 2n \):
\[ \sum_{r=1}^{2n} r = \frac{2n(2n + 1)}{2} = n(2n + 1) \]

4. **Finding the Constant \( k \)**:
Substituting the sum back into the equation for total probability:
\[ k \cdot n(2n + 1) = 1 \]
Solving for \( k \):
\[ k = \frac{1}{n(2n + 1)} \]

5. **Calculating the Probability of Drawing an Even Number**:
The even numbers from 1 to \( 2n \) are \( 2, 4, 6, \ldots, 2n \). The probability of drawing an even number can be calculated as:
\[ P(\text{even}) = P(X = 2) + P(X = 4) + \ldots + P(X = 2n) \]
Substituting the probabilities:
\[ P(\text{even}) = k \cdot 2 + k \cdot 4 + k \cdot 6 + \ldots + k \cdot 2n \]
Factoring out \( k \):
\[ P(\text{even}) = k(2 + 4 + 6 + \ldots + 2n) \]

6. **Calculating the Sum of Even Numbers**:
The sum of the first \( n \) even numbers is:
\[ 2 + 4 + 6 + \ldots + 2n = 2(1 + 2 + 3 + \ldots + n) = 2 \cdot \frac{n(n + 1)}{2} = n(n + 1) \]

7. **Substituting Back to Find \( P(\text{even}) \)**:
Now substituting back into the probability:
\[ P(\text{even}) = k \cdot n(n + 1) \]
Substituting \( k = \frac{1}{n(2n + 1)} \):
\[ P(\text{even}) = \frac{1}{n(2n + 1)} \cdot n(n + 1) = \frac{n(n + 1)}{n(2n + 1)} = \frac{n + 1}{2n + 1} \]

### Final Answer:
The probability of drawing an even number in one draw is:
\[ \frac{n + 1}{2n + 1} \]