Probability

The probability of a random phenomenon tells us how likely we are that the phenomenon will occur. A typical example would be the roll of a classic dice. We might ask "what is the probability of rolling the number five?" We give the probability either as a number from the interval <0, 1> or by using percentages, i.e. from 0% to 100%.

Related definitions

Any repeated action that depends on chance is said to be a random experiment. For example, a roll of the dice or the drawing of a sports card is a random experiment. Next, we need the set of all possible outcomes of a random experiment - that is, all the possibilities that can occur for a given random experiment. For example, numbers from one to six can occur on a die, so the set of random trials, which we label $\omega$ (omega), would look like this for a die: $\omega = {1, 2, 3, 4, 5, 6}$. For a sport, this would be all the combinations that can fall.

We call each subset of $A \subseteq \omega$ a (random) phenomenon. For example, the phenomenon "an even number rolls on the die" would be written as A = {2, 4, 6}. The phenomenon "a number greater than five rolls on the die" would be written as A = {6}, and so on. If we roll the dice (= do the experiment) and the number x rolls on the dice, then if x ∈ A, we would say that the phenomenon occurred, otherwise we would say that the phenomenon did not occur. If we stick with the "an even number rolls on the dice" phenomenon and we get x = 2, then x ∈ A, then 2 ∈ {2, 4, 6} and the phenomenon has occurred.

There are two special kinds of phenomena: certain and impossible. A certain phenomenon is if it occurs in every trial, which means that $A = \omega$. For dice, it would be the phenomenon "a number less than 7 is rolled". That's a sure thing. The opposite is an impossible phenomenon, this would be A = ∅ and verbally e.g. "a flowerpot falls on a die". Well, a flowerpot might fall off the table during a roll of the dice, but we certainly won't fall off the flowerpot on the dice because there are only numbers 1 to 6.

The probability of the phenomenon

The probability of an event tells us how much we can expect the event to occur. We probably all suspect that an even number will fall on the dice more often than the number five directly. Probability translates this hunch into exact numbers for us. The first thing we'll do is to make some assumptions.

Let the following hold for a random experiment:

• All possible outcomes are finitely many (i.e., $\omega$ is a finite set).
• Two outcomes cannot fall at the same time (i.e., we cannot roll a 3 and a 6 on the die at the same time).
• Each outcome is equally possible (i.e., we have as much chance of rolling a 4 on the die as we do of rolling a 6 or a 1).

The first two conditions are fairly natural, the last is slightly restrictive, but we'll make do for now. We can then say that the probability of the phenomenon A, denoted P(A), is equal to

$$P(A) = \frac{|A|}{|\omega|}.$$

In other words: the number of favorable outcomes divided by the number of all outcomes. (The slashes in the preceding formula denote the size of the set.) Thus the probability of rolling an even number on the die would be equal:

$$ P(Sude) = \frac{|{2{,}4,6}|}{|{1{,}2,3{,}4,5{,}6}|}=\frac36=\frac12=0{,}5. $$

We often give the probability as a percentage - just take the probability we calculated, multiply by 100, and add the percentages: 0,5 · 100 % = 50 %. The fifty percent probability actually tells us that we have as much chance of getting an even number to fall as we do of not falling. Which makes sense if we have three even numbers and three odd numbers on the die.

A certain event has probability 1, an impossible event 0. Why? We said that for a certain phenomenon J $J = \omega$ holds, and for an impossible phenomenon N N = ∅ holds. If we plug these into the formula, we get:

$$ P(J) = \frac{|\omega|}{|\omega|} = 1;\qquad P(N)=\frac{|\emptyset|}{|\omega|}=\frac{0}{|\omega|}=0. $$

Solved examples

1. What is the probability that the number five will roll on the dice? That's easy, the set of all favorable phenomena is just A = {5}, the set of all phenomena is still $\omega = {1,2,3,4,5,6}$. We plug into the formula:

   $$ P(A) = \frac{|{5}|}{|{1{,}2,3{,}4,5{,}6}|}=\frac16=16{,}666… \% $$

2. What is the probability that we roll a number divisible by three on the dice? We need to figure out which numbers on the die are divisible by three. It's only the numbers 3 and 6, so the set of admissible phenomena is B = {3, 6}. Then, classically:

   $$ P(B) = \frac{|{3, 6}|}{|{1{,}2,3{,}4,5{,}6}|}=\frac26=\frac13=33{,}333… \% $$

3. Now let's try rolling two dice at the same time. What is the probability of rolling a six on both dice? The first thing we have to do is to calculate $\omega$, i.e. the set of all phenomena. We have to calculate all the possibilities that can occur. So, if we roll a 1 on the first die, we can roll a 1 to 6 on the second die. That gives us six possibilities. If the number 2 falls on the first die, the numbers 1 to 6 can fall on the second die again; thus we have found six more possibilities. And so on for all the numbers that can fall on the first die. All the possibilities then look like this:

   $$\begin{eqnarray} &\omega=&[1{,}1], [1{,}2], [1{,}3], [1{,}4], [1{,}5], [1{,}6], \\ &&[2{,}1], [2{,}2], [2{,}3], [2{,}4], [2{,}5], [2{,}6], \\ &&\ldots\\ &&[6{,}1], [6{,}2], [6{,}3], [6{,}4], [6{,}5], [6{,}6] \end{eqnarray}$$

   We have a total of six rows, and in each row we have six possibilities. In total, we get 6 · 6 = 36 possibilities that can fall on the dice. How many of these will be in the set of favourable outcomes at the same time? Only one possibility, namely [6, 6], because we want two sixes to fall. So we get a probability:

   $$ P(C) = \frac{|{[6, 6]}|}{|\omega|}=\frac{1}{36}=2{,}777… \% $$

4. What is the probability that if we roll two dice, the same number will roll on both of them? The number of all possible outcomes is still 36, see the previous example. But how many favorable outcomes are there? In only six cases will the same number fall on both dice: D = {[1, 1], [2, 2], …, [6, 6]}:

   $$ P(D) = \frac{|D|}{|\omega|}=\frac{6}{36}=\frac16=16{,}666… \% $$

5. What is the probability that if we roll two dice, a white die and a black die, we get a 3 on the white die and a 5 on the black die? The size of $\omega$ is still 36. What does the set of favorable phenomena E look like ? It contains only the possibility [3, 5], so the probability is equal to

   $$ P(E) = \frac{|E|}{|\omega|}=\frac{1}{36}=2{,}777… \% $$

6. What is the probability that if we roll two dice, a white and a black die, the numbers 3 and 5 will be rolled? This example is similar to the previous one, except that we do not require that the number 3 falls on the white die and the number 5 falls on the black die. In short, the number 3 must fall on one of the dice and the number 5 must fall on the remaining dice. This gives us more possibilities: in addition to [3, 5] we also have the possibility of [5, 3]. The probability of this phenomenon F is thus equal:

   $$ P(F) = \frac{|F|}{|\omega|}=\frac{2}{36}=\frac{1}{18}=5{,}555… \% $$

7. What is the probability that we roll at least one 6 when we roll two dice? Let's go one at a time: if we roll a six on the first die, then the numbers 1 to 5 can fall on the second die. That's five possibilities. If a six is rolled on the second die, then numbers 1 to 5 can be rolled on the first die. That's five more possibilities. Finally, a six can roll on both dice - that's one more possibility. In total, we have 5 + 5 + 1 = 11 possibilities. So the set of allowable G phenomena looks like this: G = {[6, 1], [6, 2], [6, 3], [6, 4], [6, 5], [1, 6], [2, 6], [3, 6], [4, 6], [5, 6], [6, 6]}. We get the probability:

   $$ P(G) = \frac{|G|}{|\omega|}=\frac{11}{36}=30{,}555… \% $$

8. There are 30 students in the class. What is the probability that the pupil Andrew will be recalled in history if the mistress always tests two pupils per class and no one has been recalled yet? To calculate the probability, we need to calculate how many different pairs of pupils there are and how many different pairs of pupils there are where one of the pair is Andrew. Since the order in which the pupils are called by the mistress does not matter, we will use combinations.

   Now it's easy. We have a total of 30 pupils and we ask how many different pairs we can put together. That will lead us to a combination number

   $$ {30 \choose 2} = 435 $$

   And how many pairs does Andrew have? If one of the two is Andrew, that leaves a total of 29 pupils with whom Andrew can form a pair (he cannot form a pair with himself, so 30-1=29). Now we know the number of all possibilities and the number of favorable possibilities, so we plug them into the formula:

   $$ P(H) = \frac{29}{435}=0{,}0666\ldots=6{,}666… \% $$

   Note: the number of all pairs can be calculated without combinations. How could we form all pairs if we have a class of 30 students? By taking one pupil at a time and adding all the other pupils in turn. This means that we take Martin, for example, and add all the other pupils to his pair, i.e. we make 29 pairs. We do this with every pupil, i.e. with all 30 pupils. We get a total of 30 · 29 = 870 pairs. However, each pair is there twice, once we have the pair [Martin, Jana] and the second time [Jana, Martin] (depending on whether Martin was the one that the rest of the class was matched to or the one that was currently matched to Jana). So we divide this result by two and we have the final result: 870 / 2 = 435.

9. What is the probability of pulling out three shoes on the left foot from a shoe closet containing twelve pairs of shoes? That's such a convenient example from life, I'm sure you've needed to know. We start by counting how many different triples we can pull out of the shoe rack in the first place. We'll need combinations to do this, because it doesn't matter what order we pull the shoes out in. We have a total of 24 shoes and we want to pull out 3. We get a combination number:

   $$ {24 \choose 3} = 2024 $$

   This tells us that there are a total of 2024 different combinations of three shoes that we can pull out of the shoebox. Now we need to calculate which combinations are favorable, i.e. which combinations of three shoes always contain a shoe for the left foot. We know that there are a total of 12 shoes for the left foot in the shoebox. So we just calculate how many combinations of these 12 shoes we can make. This again leads to a combination number:

   $$ {12 \choose 3} = 220 $$

   There are 220 combinations that always contain only the left foot shoe. That's all we need to know, now we just plug it into the formula:

   $$ P(I) = \frac{220}{2024}=0{,}108695=10{,}8695 \% $$

   We have about a ten percent chance of pulling out three shoes, all for the left foot.