Probability of multiple trials? give an event $E$ and probability that event occurs as P. If it was attempted $1000$ times. What would be the answers to the following question.


*

*probability of at least the event occurring one time ? is it $1-(1-P)^{1000}$

*probability of at most the event occurring one time? is it $P*(1-P)^{999}$

*probability of at least the event occurring three times? is it $P^3*(1-(1-P)^{997})$

*probability of at most the event occurring three times ? is it $P^3*(1-P)^{997}$

*probability of event occurring more than one time ? is it  $P^2*(1-(1-P)^{998})$

*probability of event not occurring at all ? is it $(1-p)^{1000}$

 A: Hints:
The probability it happens $k$ times in $n$ independent attempts is $P(k)={n \choose k}p^k (1-p)^{n-k}$. Your attempts seem to have ignored the binomial co-efficients which count the different possible orderings of successes. 
To answer the questions, consider:


*

*$1$ minus probability $k=0$

*probability $k=0$ or $k=1$ 

*probability $k=0$ or $k=1$ or $k=2$ or $k=3$

*$1$ minus probability $k=0$ or $k=1$ or $k=2$

*$1$ minus probability $k=0$ or $k=1$

*probability $k=0$  

A: *

*probability of at least the event occurring one time ? is it $1-(1-P)^{1000}$


You don't have any other way to count the event not occurring in any other way.
The  result is $\binom{1000}{1000}$$\times${$1-(1-P)^{1000}$}


*

*probability of at most the event occurring one time? is it $P*(1-P)^{999}$


As in Lulu's comment, it has been already explained, you need to consider not occuring and only once the event occurs. Only once the event occurs, you have many other ways to count them such as the way at 1st observaton the event happens and the rest of observations the event not happens or the way at 2nd observations the event does not happen and so on so forth.
The  result is altogether not occuring and occring once. 
$\binom{1000}{1000}$$\times${$1-(1-P)^{1000}$} +  $\binom{1000}{1}$$\times${$p\times(1-P)^{999}$}


*

*probability of at least the event occurring three times? is it $P^3*(1-(1-P)^{997})$


Now You need to add another scenario the event happens twice to the last one. And you have the probability of at most twice happens. You need to remove them from all the event 1. Now you get the probability of at least three times happens.
$1$ - {$\binom{1000}{1000}$$\times${$1-(1-P)^{1000}$} +  $\binom{1000}{1}$$\times${$p\times(1-P)^{999}$} +  $\binom{1000}{2}$$\times$${p^{2}\times(1-P)^{998}}$}


*

*probability of at most the event occurring three times ? is it $P^3*(1-P)^{997}$


You need to add the probaility of three times happen to the probaility of at most twice happen.
$\binom{1000}{1000}$$\times${$1-(1-P)^{1000}$} +  $\binom{1000}{1}$$\times${$p\times(1-P)^{999}$} +  $\binom{1000}{2}$$\times$${p^{2}\times(1-P)^{998}}$ +  $\binom{1000}{3}$$\times$${p^{3}\times(1-P)^{997}}$ 


*

*probability of event occurring more than one time ? is it  $P^2*(1-(1-P)^{998})$


$1$ - { the probaility of at most the event occuring one time}


*

*the probability of event not occurring at all ? is it $(1-p)^{1000}$


Yes.
