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: $$ \prod_{i=1}^{2008} \frac{i+1}{2i+1} = \frac{2^b (b!) (c!)}{d!} $$ $\endgroup$ – Theoretical Economist Dec 26 '16 at 9:05
  • $\begingroup$ Also, obviously the upper limit of the product should be $1008$, not $2008$. $\endgroup$ – Theoretical Economist Dec 26 '16 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 '16 at 9:22

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


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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.