If the probability of choosing an integer 'k' out of 2n integers 1, 2, 3, …, 2n is inversely proportional to `k^4(1lekle2n)`. If `alpha` is the probability that chosen number is odd and `beta` is the probability that chosen number is even, then (A) `alpha gt 1/2 ` (B) `alpha gt 2/3 ` (C) `beta lt 1/2 ` (D) `beta lt 2/3 `

Options

A. `alphagt(1)/(2)`
B. `alphagt(2)/(3)`
C. `betale(1)/(2)`
D. `(betalt(2)/(3))`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understanding the Probability Function
The problem states that the probability of choosing an integer \( k \) from the integers \( 1, 2, \ldots, 2n \) is inversely proportional to \( k^4 \). This means we can express the probability \( P(k) \) as:
\[ P(k) = \frac{\lambda}{k^4} \]
where \( \lambda \) is a constant of proportionality.

### Step 2: Finding the Value of \( \lambda \)
To find \( \lambda \), we need to ensure that the total probability sums to 1:
\[ \sum_{k=1}^{2n} P(k) = 1 \]
Substituting our expression for \( P(k) \):
\[ \sum_{k=1}^{2n} \frac{\lambda}{k^4} = 1 \]
This implies:
\[ \lambda \sum_{k=1}^{2n} \frac{1}{k^4} = 1 \]
Thus, we can express \( \lambda \) as:
\[ \lambda = \frac{1}{\sum_{k=1}^{2n} \frac{1}{k^4}} \]

### Step 3: Calculating \( \alpha \) (Probability of Choosing an Odd Integer)
The odd integers in the range from \( 1 \) to \( 2n \) are \( 1, 3, 5, \ldots, 2n-1 \). There are \( n \) odd integers. The probability \( \alpha \) can be calculated as:
\[ \alpha = \sum_{k=1}^{n} P(2k-1) = \sum_{k=1}^{n} \frac{\lambda}{(2k-1)^4} \]
Substituting for \( \lambda \):
\[ \alpha = \frac{1}{\sum_{k=1}^{2n} \frac{1}{k^4}} \sum_{k=1}^{n} \frac{1}{(2k-1)^4} \]

### Step 4: Calculating \( \beta \) (Probability of Choosing an Even Integer)
The even integers in the range from \( 1 \) to \( 2n \) are \( 2, 4, 6, \ldots, 2n \). There are also \( n \) even integers. The probability \( \beta \) can be calculated as:
\[ \beta = \sum_{k=1}^{n} P(2k) = \sum_{k=1}^{n} \frac{\lambda}{(2k)^4} = \sum_{k=1}^{n} \frac{\lambda}{16k^4} \]
Substituting for \( \lambda \):
\[ \beta = \frac{1}{\sum_{k=1}^{2n} \frac{1}{k^4}} \sum_{k=1}^{n} \frac{1}{16k^4} \]

### Step 5: Comparing \( \alpha \) and \( \beta \)
Since \( \alpha \) involves summing over odd integers and \( \beta \) involves summing over even integers, we can analyze their relationship.

Given that:
\[ \alpha + \beta = 1 \]
If we show that \( \alpha > \beta \), it follows that:
\[ \alpha > \frac{1}{2} \quad \text{and} \quad \beta < \frac{1}{2} \]

### Conclusion
From our findings:
- \( \alpha > \frac{1}{2} \)
- \( \beta < \frac{1}{2} \)

Thus, the correct options are:
- (A) \( \alpha > \frac{1}{2} \)
- (C) \( \beta < \frac{1}{2} \)