Probability Calculation: Independent Events

The problem asks for the probability that none of the given independent events $A_1, A_2, ..., A_{1006}$ occurs. This probability is calculated as the product of the probabilities of the complements of these events.

Step 1: Complement Probabilities

For each event $A_i$, the probability of its complement $A_i^c$ occurring is:

$P(A_i^c) = 1 - P(A_i) = 1 - \frac{1}{2i} = \frac{2i-1}{2i}$

Step 2: Product of Complement Probabilities

Since the events $A_i$ are independent, their complements $A_i^c$ are also independent. Therefore, the probability that none of the events occurs is the product:

$P(\text{none occurs}) = \prod_{i=1}^{1006} P(A_i^c) = \prod_{i=1}^{1006} \left( \frac{2i-1}{2i} \right)$

Expanding this product:

$P(\text{none occurs}) = \frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \times ... \times \frac{2(1006)-1}{2(1006)}$

$P(\text{none occurs}) = \frac{1 \cdot 3 \cdot 5 \cdot ... \cdot 2011}{2 \cdot 4 \cdot 6 \cdot ... \cdot 2012}$

Step 3: Factorial Representation

To express this product in terms of factorials, we use the identity for the product of the first $n$ odd numbers: $1 \cdot 3 \cdot ... \cdot (2n-1) = \frac{(2n)!}{2^n n!}$. The product of the first $n$ even numbers is $2 \cdot 4 \cdot ... \cdot (2n) = 2^n n!$.

In this case, $2n = 2012$, which means $n = 1006$.

Numerator: $1 \cdot 3 \cdot ... \cdot 2011 = \frac{2012!}{2^{1006} 1006!}$

Denominator: $2 \cdot 4 \cdot ... \cdot 2012 = 2^{1006} 1006!$

Thus, the probability is:

$P(\text{none occurs}) = \frac{\frac{2012!}{2^{1006} 1006!}}{2^{1006} 1006!} = \frac{2012!}{2^{2012} (1006!)^2}$

Step 4: Identify $\alpha$ and $\beta$

The problem states that this probability is equal to $\frac{\alpha!}{2^\alpha (\beta!)^2}$. By comparing the derived expression with the given form:

• $\alpha! = 2012! \implies \alpha = 2012$
• $2^\alpha = 2^{2012} \implies \alpha = 2012$
• $(\beta!)^2 = (1006!)^2 \implies \beta = 1006$

So, we have $\alpha = 2012$ and $\beta = 1006$.

Step 5: Option Verification

We now check the given options using $\alpha = 2012$ and $\beta = 1006$.

• Option B: $\alpha = 2\beta$ Substituting the values: $2012 = 2 \times 1006$. This statement is true.
• Option C: $\beta$ is of the form $4k+1, k \in I$ The problem implies that $\beta$ must satisfy this condition. Therefore, we select this option.