If `A_1,A_2,A_3,........A_1006` be independent events such that `P(A)=1/(2i)(i=1,2,3,.....1006)` and probability that none of the events occur be `(alpha!)/(2^alpha(beta!)^2).` then

Options

A. `beta` is of form `4k+2, k in I`
B. `alpha=2beta`
C. `beta` is a composite number
D. `alpha` is of form `4k, k in I`

Answer

To solve the problem step by step, we will analyze the given information and derive the necessary conclusions.

### Step 1: Understanding the probabilities of the events
Given that the probabilities of the independent events \( A_1, A_2, A_3, \ldots, A_{1006} \) are:
\[ P(A_i) = \frac{1}{2i} \quad \text{for } i = 1, 2, 3, \ldots, 1006 \]

### Step 2: Finding the probability that none of the events occur
The probability that none of the events occur is given by:
\[ P(A_1') \cdot P(A_2') \cdot P(A_3') \cdots P(A_{1006}') \]
where \( P(A_i') = 1 - P(A_i) = 1 - \frac{1}{2i} = \frac{2i - 1}{2i} \).

Thus, we can express the probability that none of the events occur as:
\[ P(A_1') \cdot P(A_2') \cdots P(A_{1006}') = \prod_{i=1}^{1006} \frac{2i - 1}{2i} \]

### Step 3: Simplifying the product
This product can be simplified as follows:
\[ P(A_1') \cdot P(A_2') \cdots P(A_{1006}') = \frac{1 \cdot 3 \cdot 5 \cdots \cdot (2 \cdot 1006 - 1)}{2^{1006} \cdot (1 \cdot 2 \cdot 3 \cdots \cdot 1006)} \]
The numerator is the product of the first 1006 odd numbers, which can be expressed as:
\[ \frac{(2 \cdot 1006)!}{2^{1006} \cdot 1006!} \]
Thus, we have:
\[ P(A_1') \cdot P(A_2') \cdots P(A_{1006}') = \frac{(2 \cdot 1006)!}{2^{1006} \cdot 1006! \cdot 2^{1006} \cdot 1006!} = \frac{(2 \cdot 1006)!}{2^{2012} \cdot (1006!)^2} \]

### Step 4: Relating to given expression
We know from the problem statement that this probability can also be expressed as:
\[ \frac{\alpha!}{2^\alpha (\beta!)^2} \]
By comparing the two expressions, we can identify:
\[ \alpha = 2012 \quad \text{and} \quad \beta = 1006 \]

### Step 5: Verifying conditions
1. **Check if \( \beta \) is of the form \( 4k + 2 \)**:
\[ \beta = 1006 = 4 \cdot 251 + 2 \quad \text{(which is of the form \( 4k + 2 \))} \]

2. **Check if \( \alpha = 2\beta \)**:
\[ \alpha = 2012 = 2 \cdot 1006 \quad \text{(which is true)} \]

3. **Check if \( \beta \) is a composite number**:
\[ \beta = 1006 \quad \text{(which is divisible by 2, hence composite)} \]

4. **Check if \( \alpha \) is of the form \( 4k \)**:
\[ \alpha = 2012 = 4 \cdot 503 \quad \text{(which is of the form \( 4k \))} \]

### Final Conclusion
Thus, we have verified all the conditions stated in the problem.