# Definition of probability, sample space and sure and impossible event

## Definition of probability

"It may rain tomorrow", "When rolling a dice it is more probable we get a number greater than four than a one", "Probably that issue will be in the exam", "It is little probable you win the lottery."

We all have an intuitive notion on probability, but what is it exactly? Although we have wondered how chance works for centuries, or whether it is possible to predict the future, it was not until the sixteenth century with the work of Cardano and Tartaglia, that the resolution of these mathematical problems began to progress.

Probability, as we understand it today, was born in the seventeenth century, when Pierre de Fermat and Blaise Pascal were sending each other a series of letters by which they were trying to solve problems about gambling. In these letters, they were trying to find methods and a mathematical notation to solve problems related to probability.

We can consider what probability is with the following example.

We take a dice and we write down how many fours appear when we throw it $5, 10, 20, 50$ and $100$ times. Let's suppose that we get something like:

 Throws Number of fours $5$ $2$ $10$ $3$ $20$ $5$ $50$ $8$ $100$ $17$

Now let's consider the proportion of fours regarding the total amount of throws:

$$\dfrac{2}{5}=0'4, \ \dfrac{3}{10}=0'3, \ \dfrac{5}{20}=0'25, \ \dfrac{8}{50}=0'16, \ \dfrac{17}{100}=0'17$$

After this experiment we may wonder: "if I throw the dice again: what is the probability of getting a four?"

It is true that the final result will depend on chance, but we have seen that if we throw the dice many times, it is normal to get a four about $17$ times out of $100$. Therefore, we say that the probability is approximately $17\%$, or in other words, $0'17$.

In fact, if we think more deeply about it, as a dice has six faces, and all have the same probability to show, we can expect that in $6$ throws one will be a four, that is, we think that the probability should be $$\dfrac{1}{6}=0'1\widehat{6}=0'1666\ldots$$

This will be the basis of the law of Laplace.

## Sample space and events

A problem begins: "We threw a dice ..." or "We threw a coin ..." What do we do? First, we have to now what results may come out:

In the case of the coin, it can be ''heads''($h$) or ''tails''($t$).

If we throw a dice, "$1$", "$2$", "$3$", "$4$", "$5$" or "$6$." These are the possible outcomes, also called elementary events.

Of course, the elementary events are different in each problem. If you throw a coin twice, then the outcomes can be "$hh$", "$ht$" "$th$", "$tt$": first the outcome of the first throw, and next the outcome of the second one.

We can even think that our experiment is to go out and watch the first man we find: our elementary events will be if he has a "beard", "beard and moustache," "only a moustache," or if he is "clean-shaven."

The set of all the possible outcomes is called sample space, and it is usually represented with the Greek letter $\Omega$.

In this way, in the four previous examples, the sample space would be:

$\Omega=\lbrace \mbox{head,tail} \rbrace = \lbrace h,t \rbrace$

$\Omega=\lbrace 1,2,3,4,5,6 \rbrace$

$\Omega=\lbrace (h,h), \ (h,t), \ (t,h), \ (t,t) \rbrace$, to simplify we can write $\Omega=\lbrace hh, \ ht, \ th, \ tt\rbrace$

$\Omega=\lbrace \mbox{beard, beard and mustache, mustache, shaved} \rbrace$

We continue now with the statement. "We threw a dice. What is the probability of getting a four?"

What we have to consider, in this case that "it comes a four," is what we call events, which are subsets of the sample space.

Thus, in this case our sample space is $\Omega= \lbrace 1,2,3,4,5,6 \rbrace$, and our success is $A=\mbox{"get a four"}=\lbrace 4 \rbrace$.

Also we might have considered other events, for instance, getting an even number. Then, our event would be $B=\mbox{"get an even number"}=\lbrace 2,4,6 \rbrace$. If the event is an "extract three or five", then $C=\mbox{"get a 3 or a 5"}=\lbrace 3,5 \rbrace$.

We can think of events as the results we want to consider among all possible outcomes, that is, from the total sample space.

## Sure event and impossible event

A sure event is the one that contains the whole sample space. For example, in our experiment of throwing a dice and noting the result, the event $A=\mbox{"get a number smaller than} \ 2 \mbox{ or higher than or equal to } 3"$. Notice that $A=\lbrace 1,2,3,4,5,6 \rbrace = \Omega$.

There will be cases in which it is not clear if it is a sure event.

For example, suppose that we toss two coins, and want to know if the event $C="\mbox{get head in one of the two coins, or get the same result in both}"$ is sure.

First: we write the sample space. $\Omega=\lbrace hh, \ ht, \ th, \ tt\rbrace$

Second: we write our event. What elementary events form the event?

If we think a little we can see that it is, in fact, a safe event because, if we get "$hh$" or "$tt$", with both coins we get the same, and in the other two possible cases, "$h t$" and "$t h$", we get a head in one of the coins. That is, no matter the result we get, $C$ will always happen.

An impossible event is the opposite case, when the event does not contain any element of the sample space.

For example, the event $A="\mbox{get a } 7"$ on throwing a dice of six faces, or $B="\mbox{get a white ball}"$ from an urn that only contains black balls.

Usually, these events are represented with $A=B=\emptyset$, which is the empty set, or in other words, we are saying that there is no possible result that satisfies the event.

Let's see some examples:

We have three urns, and each one has white and black balls. We draw a ball of each urn, and look at its color.

1. Describe the sample space of the experiment.
2. Describe the event $A=$ "draw only a white ball"

1

We will consider that $W=$"white ball", $B=$"black ball".

Since we extract a ball of each urn, every elementary event of our experiment has three balls, which can be white or black. That is, $\Omega=\lbrace WWW,WWB,WBW,WBB,BWW,BWB,BBW,BBB \rbrace$, where "WWB" means: "get a white ball in the first urn, white in the second urn and black in the third urn".

Depending on what we were interested in calculating, it would not be incorrect to think that our elementary events are mixed up, and the only important thing is the total number of white and black balls that we have extracted. In this case, our sample space would be $\Omega'=\lbrace WWW,WWB,WBB,BBB \rbrace$, where, now "WWB" just means: "get two white and one black balls", but usually we will not use this way of calculating, for it may lead to problems when counting probabilities, as we will see.

2

We can draw a white ball in any of the three urns. Therefore, all the events that we are interested in are: $WBB, BWB, BBW$. Namely, $A= \lbrace WBB, BWB, BBW \rbrace$.

We throw a dice with eight faces. If we get a one, we throw the dice only once more, and we add up the results.

1. What is the sample space of this experiment?
2. Describe the events $A =$ "get a number smaller than $2$", $B =$ "get a five or six", $C =$ "get a multiple of three"

1

When we throw a dice of eight faces, the results that can be $\lbrace 1,2,3,4,5,6,7,8 \rbrace$.

However, in our experiment, when we get a one, we throw the dice again, and we can obtain again a number between $1$ and $8$.

The smallest result that we can obtain is $2$ (if we get a $1$ the first time, and another $1$ the second time), and the biggest result is $9$ if we get a $1$ the first time, and $8$ the second time). Therefore, our sample space is $\Omega=\lbrace 2,3,4,5,6,7,8,9 \rbrace$.

Note that "$1$" is not an elementary event of our experiment, since we will never be able to obtain one as result by adding two throws together.

2

We cannot obtain a result less than $2$, so that $A=\emptyset$. That is, $A$ is an impossible event.

The events that satisfy $B$ are $5$ and $6$ since if we get any of those, the event is satisfied. Therefore, $B=\lbrace 5,6 \rbrace$.

To calculate $C$, we have to look at which are the multiples of three of our sample space. In our case, we have $3,6,9$. Therefore, $C=\lbrace 3,6,9 \rbrace$.