If $E_1$,$E_2$,$E_3$.....$E_{1008}$ be $1008$ independent events such that $P(E_i)=\frac{i}{2i+1}$;$(i=1,2,3.....1008)$ and probability that none of the events occur be $\frac{2^b(b!)(c!)}{(d!)}$ where $b,c,d$ are natural numbers such that $b<c<d$.Then what is the relation between $b,c,d$ also if possible what are their values?

My Try:I know that probability in case of independent events is the product of individual probabilities of all the independent events. But I can't build up on this. Can someone please tell me how to proceed or how to get the question done?

  • $\begingroup$ $$P(E_i^c)=1-\frac{i}{2i+1} = \frac{i+1}{2i+1}$$ Hence: $$ \prod_{i=1}^{2008} \frac{i+1}{2i+1} = \frac{2^b (b!) (c!)}{d!} $$ $\endgroup$ Dec 26, 2016 at 9:05
  • $\begingroup$ Also, obviously the upper limit of the product should be $1008$, not $2008$. $\endgroup$ Dec 26, 2016 at 9:18
  • $\begingroup$ I already reached till $$ \prod_{i=1}^{1008} \frac{i+1}{2i+1} = \frac{2^b (b!) (c!)}{d!} $$ can someone help me solve further? I don't know how to solve the product function. $\endgroup$
    – Zlatan
    Dec 26, 2016 at 9:22

1 Answer 1


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


$$ \prod_{i=1}^{1008} \frac{i+1}{2i+1} = \frac{1009!}{2017\cdot 2015 \cdot 2013 \cdots 3 \cdot 1}$$


It would be correct to guess that $1009!$ corresponds to one of the factors appearing in the numerator of $$ \frac{2^b(b!)(c!)}{d!} $$

After making that guess, the two other variables are easy to determine.


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