Question

If $A_1,A_2,A_3,......A_{1006}$ are independent events such that $P\left(A_i\right)=\frac{1}{2i}$, where $i=1,2,\mathrm{3.....1006}$ and the probability that none of the events occurs is $\frac{\alpha !}{2^{\alpha }(\beta !{)}^2}$, then the value of $\alpha +\beta$ is

Answer 1

Given : $P\left(A_i\right)=\frac{1}{2i}$

For the none of the events to occur
$P\left(\text{None of the event occurs}\right)=\prod _{i=1}^{i=1006}(1-\frac{1}{2i})=\prod _{i=1}^{i=1006}\left(\frac{2i-1}{2i}\right)=\frac{1}{2^{1006}}\prod _{i=1}^{i=1006}\left(\frac{2i-1}{i}\right)=\frac{1\cdot 3\cdot 5\cdot 7\cdots \cdots 2011}{2^{1006}\cdot 1006!}$

Multiplying by $(2\cdot 4\cdot 6\cdot 8\cdots \cdots 2012)$ on both denominator and numerator

$=\frac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdots \cdots 2011\cdot 2012}{2^{1006}\cdot 1006!(2\cdot 4\cdot 6\cdot 8\cdots \cdots 2012)}=\frac{2012!}{2^{2012}\cdot (1006!{)}^2}$

Therefore comparing with
$\frac{\alpha !}{2^{\alpha }(\beta !{)}^2}$
We get
$\alpha =2012,\beta =1006\therefore \alpha +\beta =3018$