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 ?

  • $\begingroup$ Perhaps they meant that if you have $N$ independent events (in one-dimensional probability space) with $k$ outcomes (for coin tossing $k=2$, for common dice $k=6$, for roulette $k=37$ or $38$ depending on country) and repeat the experiment $N$ times, then you will have exactly $k^N$ events (in $N-$dimensional probability space) in total. This is not working for some dependent one-dimensional events when elements are removed after they are selected, so you will have less than $k^N$ possible events in related probability spaces. $\endgroup$ – Andrei Rykhalski Feb 20 '15 at 12:20

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.

  • $\begingroup$ But its a case of one toss as per the definition in my question. i got it from university of washington notes. So, if we are tossing only once, how come "a head and tail is thrown". I suppose you mean "and" as getting both outcomes. $\endgroup$ – user3116355 Feb 20 '15 at 12:24
  • $\begingroup$ Not very likely indeed to throw a head and a tail by one toss. That is exactly the reason why it has probability $0$. The theory somehow says: we don't exclude it, but just give it this probability $0$ to repair that. $\endgroup$ – drhab Feb 20 '15 at 12:27
  • $\begingroup$ ok. I got a slight idea. It may happen or may not. If it happens, it has 2 outcomes(either head or tail). So, in the probabilistic perspective, should we consider the chances of whether it happens or not. How can we consider it as outcomes? Outcomes occur if experiement is successful. I am rookie in the area. So, if I am talking rubbish, please dont mind. $\endgroup$ – user3116355 Feb 20 '15 at 12:30
  • $\begingroup$ By an experiment (throwing a toss or a die) we must built an appropriate mathematical model and start with a set of outcomes. Based on those outcomes we have events as sets of outcomes. The event "occurs" if the outcome lands in the set that is linked to that event. $\endgroup$ – drhab Feb 20 '15 at 12:39

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)
  • $\begingroup$ I am not sure why ppl downvoting $\endgroup$ – Mina Gabriel Apr 24 '18 at 22:33
  • $\begingroup$ The downvotes may be because drhab's answer is correct -- he just refers to the same above four events as "a head is thrown" $(\{H\}),$ "a tail is thrown" $(\{T\}),$ "a head and a tail is thrown" $(\{H\}\cap\{T\}=\{\}=\emptyset,$ and "a head or a tail is thrown" $(\{H\}\cup\{T\}=\{H,T\}=\Omega.$ $\endgroup$ – r.e.s. Jul 14 at 5:06

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"

  • $\begingroup$ So you are telling, if i throw a dice I can get a even or odd ( as an example case ). I thought the same way initailly. But what hits me was, to take the chance of possibility of getting odd number say (1,3,5), we have to throw the dice 3 times. But the definition in the question holds for number of possible outcomes. they are not mentioning the number of times we have to perform an experiment. $\endgroup$ – user3116355 Feb 20 '15 at 12:27
  • $\begingroup$ An event is different from an outcome. The event is "getting an odd number", so any number in {1,3,5} is in the event (you can say "realise the event") $\endgroup$ – Tryss Feb 20 '15 at 12:34
  • $\begingroup$ @user3116355 You can get an odd number by throwing the die only once. $\endgroup$ – drhab Feb 20 '15 at 12:34
  • $\begingroup$ @Tryss - I got it. Suppose we are having three outcomes say {a,b,c}. So, N = 3. Events E = 2^N=2^3 =8. So, events is the subset of outcomes. So we will have: [{a,b,c},{a},{b},{c},{a,b},{a,c},{b,c},{Nullset}]. So total 8 subsets. $\endgroup$ – user3116355 Feb 20 '15 at 12:47
  • $\begingroup$ @drhab - Please check the answer above. $\endgroup$ – user3116355 Feb 20 '15 at 12:48

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 $2^{|\Omega|}$.

  • $\begingroup$ "has to be" and "basically" are a bit too strong. However, within this context it is okay. $\endgroup$ – drhab Feb 20 '15 at 12:55
  • $\begingroup$ Good point. We can construct the probability measure in a number of ways deciding which subsets are measurable or not. For a discrete random variable with $n$ outcomes such as the example here, it will always be exactly $2^n$, yes? Of course many of those events could have measure zero. $\endgroup$ – jdods Feb 20 '15 at 13:51

*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)].


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 $2^N.$


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