Question

If \(A_{1},A_{2},A_{3},\dots,A_{1006}\) be independent events such that \[ P(A_{i})=\frac{1}{2i}\quad (i=1,2,\dots,1006) \] and the probability that none of the events occurs be \[ \frac{\alpha!}{2^{\alpha}(\beta!)^{2}}, \] then:

• A. $\beta$ is of the form $4k+2, k\in I$
• B. $\alpha=2\beta$
• C. $\beta$ is of the form $4k+1, k\in I$
• D. $\beta$ is a prime number

Hint:

To turn a product of consecutive odd numbers into a factorial quickly, use the standard double-factorial identity: $(2n-1)!! = \frac{(2n)!}{2^n \cdot n!}$. Squaring the denominator to account for the even terms instantly reveals the values of $\alpha$ and $\beta$ without manual expansion.

Answer 1

The Correct Option is A

Concept: For a collection of independent events, the probability that none of them occur is equal to the product of the probabilities of their individual complementary events: $P(\text{None}) = \prod (1 - P(A_i))$. Step 1: Write out the product sequence for the complementary events.
We are given that the probability of each individual event is $P(A_i) = \frac{1}{2i}$. The probability of its complement (the event not occurring) is: \[ P(A_i') = 1 - \frac{1}{2i} = \frac{2i - 1}{2i} \] Since all the events are independent, the total combined probability that none of them occur is the product of these complements from $i = 1$ to $n = 1006$: \[ P(\text{None}) = \prod_{i=1}^{1006} \frac{2i - 1}{2i} = \frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \times \dots \times \frac{2011}{2012} \quad \cdots (1) \]

Step 2: Transform the product into a factorial expression.
To convert this product of odd numbers into standard factorials, let us multiply both the numerator and the denominator by the product of all the missing even numbers: $(2 \times 4 \times 6 \times \dots \times 2012)$: \[ P(\text{None}) = \frac{(1 \times 2 \times 3 \times 4 \times \dots \times 2011 \times 2012)}{(2 \times 4 \times 6 \times \dots \times 2012)^2} \] The numerator is now a straightforward factorial: $2012!$. Let us simplify the denominator by factoring out a 2 from each of the 1006 individual even terms: \[ (2 \times 4 \times 6 \times \dots \times 2012) = 2^{1006} \cdot (1 \times 2 \times 3 \times \dots \times 1006) = 2^{1006} \cdot (1006!) \] Substitute this back into the denominator slot: \[ P(\text{None}) = \frac{2012!}{\left(2^{1006} \cdot 1006!\right)^2} = \frac{2012!}{2^{2012} \cdot (1006!)^2} \quad \cdots (2) \]

Step 3: Extract the values of the tracking parameters $\alpha$ and $\beta$.
Compare our derived factorial expression from equation (2) directly with the template format given in the problem statement, $\frac{\alpha!}{2^{\alpha}{(\beta!)}^{2}}$: \[ \alpha = 2012 \quad \text{and} \quad \beta = 1006 \]

Step 4: Verify the properties of the parameters to check the options.
Let us test the relationship between our extracted values against the answer options:
• Checking Option (B): Check the ratio link: $2\beta = 2(1006) = 2012 = \alpha \implies \alpha = 2\beta$. This confirms that option (B) is correct.
• Checking Options (A) and (C): Let us divide $\beta = 1006$ by 4 to check its remainder structure: \[ 1006 = 4(251) + 2 \quad \Rightarrow \quad \beta = 4k + 2 \] This matches the remainder template for option (A) perfectly, confirming it is correct and filtering out option (C). Therefore, the correct options are (A) and (B).