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

Step 1: Understanding the Concept:
For independent events, the probability that none occur is the product of the probabilities of their complements: \( P(\text{None}) = \prod (1 - P(A_i)) \). We then manipulate the product of fractions to identify factorials.
Step 2: Key Formula or Approach:
1. \( P(\text{None}) = \prod_{i=1}^{1006} (1 - \frac{1}{2i}) \).
2. Multiply and divide by missing even terms to create a full factorial.
Step 3: Detailed Explanation:
\[ P(\text{None}) = \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \dots \cdot \frac{2011}{2012} \]
To form a factorial in the numerator, multiply numerator and denominator by \( 2 \cdot 4 \cdot 6 \dots 2012 \):
\[ \text{Num} = (1 \cdot 3 \dots 2011) \times (2 \cdot 4 \dots 2012) = 2012! \]
\[ \text{Den} = (2 \cdot 4 \cdot 6 \dots 2012)^2 = (2^{1006} \cdot 1006!)^2 \]
\[ P(\text{None}) = \frac{2012!}{2^{2012} \cdot (1006!)^2} \]
Comparing with the form \( \frac{\alpha!}{2^\alpha(\beta!)^2} \):
\( \alpha = 2012 \) and \( \beta = 1006 \).
Check the relationship: \( \alpha = 2012 = 2 \times 1006 = 2\beta \). So (B) is correct.
Check the form of \( \beta \):
\( \beta = 1006 \). Divide by 4: \( 1006 = 4 \times 251 + 2 \).
So \( \beta \) is of the form \( 4k + 2 \). So (A) is correct.
Step 4: Final Answer:
Options (A) and (B) are correct.