Number of possible events is $2^N$ ( $N$ is the number of outcomes )? A probability space is a random process or experiment with
three components:
–
$Ω,$ the set of possible outcomes $O$
*number of possible outcomes = $|Ω| = N$
*$F$, the set of possible events $E$
      - an event comprises $0$ to
$N$ outcomes


*

*number of possible events
= $| F | = 2^N$ 


Here I am not able to extract the exact meaning of number of possible outcomes. In case of tossing a coin, we have only $2$ outcomes. So, $N = 2$. So, either we have head or tail. According to the above definition, $|F| = 2^N = 2^2 = 4.$ How come this formula or axiom is valid in probability space ?
 A: The answer by drhabs is wrong, his answer misses the null set. 2^n comes from the fact that for each element in your set you have 2 choices, to either include or not include it in a subset. For the example Ω={H,T} you will have the following subsets:


*

*{H,T} (every set is a subset of it's self)

*{H} 

*{T} 

*{} (null set)

A: In the situation you sketch we have $\Omega=\{H,T\}$ where $H$ stands for head and $T$ for tail. 
There are $2$ possible outcomes and there are $2^2=4$ events:


*

*a head is thrown

*a tail is thrown

*a head and a tail is thrown (has probability $0$)

*a head or a tail is thrown (has probability $1$)


I am not sure whether this answers your question, but I hope this makes things more clear for you.
A: an event is different from an outcome.
You have also the event "the coin land" $= E$ and the event "the coin doesn't land" $= \emptyset$
Take a dice, it's more clear : You can roll 1 to 6, but an event can be "roll an even number" or "roll a number under 5"
A: The set of possible events $F$ has to be the collection of all sets which can be constructed from the individual outcomes using the set operations of union and intersection. $F$ is basically the set of all subsets of $\Omega$, so $F$ has size $2^{|\Omega|}$.
A: *number of possible outcomes = |Ω|=N
The above statement should be the following
number of possible outcomes "in an experiment" = |Ω|=N
Then, everything would be clear.
In the case of tossing a coin, if you do two times then N = 2.
Sample space is equal 2^2 = 4.
The elements would be [(H,T),(T,H),(T,T),(H,H)]. 
A: Lets first see what an event means.
An event is not one outcome, but is a set.
For example in a problem of tossing a coin, the outcomes are $H$ & $T.$ 
So the outcome set (Lets say $O$) , $O = \{H,T\}.$
Now an event is a subset of $O$, which means that it will contain the nullset, each single outcome and their combinations. i.e $E$ (all possible events) = $\{0,H,T,\{H,T\}\}.$
To understand why the number of events is equal to $2^N$ , lets take a outcome set $O$ having $N$ elements.
Null set can be selected in $1$ way
A event with one out the $N$ outcomes will be selected in $N$ ways,
with $2$ of the outcomes in NC2 ways and so on
when you add all of them,
its the binomial expansion of $2^N.$
A: $2^N$ is very special. It describes so much and is a key to understanding something truly magical.
An event that can only have $2$ outcomes and more so, can occur equally likely, event conducted in a fair setting. Fair as in pure, without any bias or favouring either of the outcome deliberately! Conducting a trial, Bernulli trial, determining an answer, detecting level of responsiveness.
Fundamentally, during the occurrence of the most simplest event possible, there are only $2$ possible states for this event to be in. At any given time of conducting a measurement the event has not yet occured or it already has.
Consider a sequence, of question -> answer pairs:
Has the event occured? No
Has the event occured? No
Has the event occured? Yes
Has the event occured? Yes
Has the event occured? No
In such case the ratio of number of No's and Yes's observed will be equal
Any sequnce of events with $2$ possible states of outcome has a chance of being observed once in $2^N$ number of trials.
$$\frac12 \times \frac12 \times \ldots = \frac1{2^N}$$
Knowing this it is possible to "game the system".
Here's the simplest thought experiment illustrating the situation if this exploitation occurring. In a way one can use randomness against itself like this:
Decide on the outcome in advance (Red) and conduct the experiment (Throw the ball).
Has the ball stopped on Red? No
Has the ball stopped on Red? No
Has the ball stopped on Red? No
Has the ball stopped on Red? No
Has the ball stopped on Red? Yes
Bet $1$ is $\$1$. Loss. Total loss is $\$1$
Bet $2$ is $\$2$. Loss. Total loss $1+2=3$
Bet $3$ is $\$4$. Loss. Total loss $+4=7$
Bet $4$ is $\$8$. Loss. Total loss $+8=15$
Bet $5$ is $\$16$. Win. Total win is equal to my original $\$16$ used to place bet 5 coming back, plus the win itself. Win equal to the amount I placed on 5th coin toss and $16 additional. Bringing my total win to $32=$16+$16
Lets count our chickens.
Our total loss was $\$15$ the casino took it and that money is not coming back. $-\$15$.
Win is $+\$32$. Assume original $\$16$ were borrowed from a friend to play in casino. So we return his money back $-\$16$ and keep the other earned $\$16$ to ourselves. We've spent $\$15$ out of our pocket to play the coin toss game and ended up with $\$16$ so our net win $-\$15+\$16=\$1$.
Introducing a particular notation O(n)=1 means even if you throw n number of tosses and lose -$X amount and win back +$Y amount you'll end up with -$X+$Y=$1 always at the end.
Next time you throw the coin 4 times and in that circumstance lose 3x but win on the 4th you'll lose -$7 and win +$8 you'll end up now with $2 extra.
Here's what I mean:
Consider a sequence of 5+3 coin tosses:
[Loss Loss Loss Loss Win Loss Loss Win]  or simply [0 0 0 0 1 0 0 1] shows after observing 2 wins you'll end up with $2 extra.
So again O(n)=1 means tossing the coin n times so long as a win at the end has occured will always bring $1 extra.
I keep increasing my bet at the rate of 2 times the previously placed bet so eventually a win is observed and I end up with an extra $1 or an extra unit if your initial bet was not $1 but something else.
The next bet after a win (6th flip or 9th flip) will be back to $1 since we reset the event and start anew.
2^N-1 is also an important formula. It is the total sum, the closed form solution to the SUM(2^N). At any given point in time the total loss is 2^N-1. The sequence of N losses is equal to $2^N-1 dollars.
What if we don't double?
Bet $1$ is $\$1$. Loss. Total loss is $-\$1$
Bet $2$ is $\$3$. Loss. Total loss $-4$
Bet $3$ is $\$7$. Loss. Total loss $-\$11$
Bet $4$ is $\$15$. Loss. Total loss $-26$
Bet $5$ is $\$31$. Win. Total win is $2\times\$31=62$. Net profit $\$-(26+31)+2\times31=+\$5$
Number of flips is $5$. Profit for this batch of flips is $5$.
Bet $6$ is $\$1$. Loss. Total loss is $-1$
Bet $7$ is $\$3$. Loss. Total loss $-4$
Bet $8$ is $\$7$. Win. Total loss is $-11$, total win is $\$14$ so net profit is $-11+14=3$.
Number of flips is $3$. Total profit is $3$ for the second batch of flips.
Total profit is $5+3=8$. Total number of flips is $8$.
Using the introduced notation above $O(n)=n$.
This would mean flip a coin $5$ times and bet $\$100$ on the first flip, the total profit will be $\$500$. Flip another $3$ times, another $\$300$. So long as you can sustain your bets that grow exponentially once a win is observed total number will always be equal to n. So $f(x)=x$.
Now think about $\mathcal{O}(n)=n^2$
Here flip a coin $5$ times and win $\$2500$ plus $\$900$ if you flip another $3$ times.
