Probability

The probability of a random event tells us how likely it is that the event happens. A typical example is rolling an ordinary dice. We might ask, “what is the probability that we roll a five?” We give probability either as a number from the interval [0, 1] or as a percentage, from 0% to 100%.

Related definitions

Any repeatable activity that depends on chance is called a probability experiment. For example, rolling a dice or the National Lottery draw is a probability experiment. We also need the set of all possible outcomes of the experiment – these are all the things that can happen in a given experiment. For example, a dice can show the numbers from one to six, so the set of all possible outcomes, called the sample space and written ω (omega), would look like this for a dice: ω = {1, 2, 3, 4, 5, 6}. For a lottery it would be all the combinations that could be drawn.

Every subset A ⊆ ω is called an event. For example, the event “the dice shows an even number” would be written as A = {2, 4, 6}. The event “the dice shows a number greater than five” would be written as A = {6}, and so on. If we roll the dice (= carry out the experiment) and the dice shows the number x, then if x ∈ A we say that the event has happened, and otherwise we say that it has not happened. If we stay with the event “the dice shows an even number” and we roll x = 2, then x ∈ A, that is, 2 ∈ {2, 4, 6}, and the event has happened.

There are two special kinds of event: certain and impossible. An event is certain if it happens in every experiment, which means that A = ω. For a dice this would be the event “the number rolled is less than 7”. That will definitely happen. The opposite is the impossible event, which would be written A = ∅ and described in words as, for example, “the dice shows a flowerpot”. Well, a flowerpot might fall off the table while we are rolling the dice, but a flowerpot will certainly not appear on the dice, because it only has the numbers 1 to 6 on it.

The probability of an event

The probability of an event tells us how much we can expect the event to happen. We all probably sense that an even number comes up on a dice more often than a five specifically. Probability turns this feeling into exact numbers. First we state some assumptions that we will need.

Suppose that for a probability experiment the following hold:

• There are only finitely many possible outcomes (that is, ω is a finite set).
• Two outcomes cannot happen at the same time (that is, we cannot roll a 3 and a 6 on a dice at once).
• Every outcome is equally likely (that is, we have the same chance of rolling a 4 on a dice as of rolling a 6 or a 1).

The first two conditions are fairly natural, and the last is slightly restrictive, but we will manage with it for now. Then we can say that the probability of an event A, which we write P(A), is equal to

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

In other words: the number of favourable outcomes divided by the number of all outcomes. (The vertical bars in the formula above denote the size of a set.) The probability of rolling an even number on a dice would therefore be:

$$P(\text{even})=\frac{|\{2,4,6\}|}{|\{1,2,3,4,5,6\}|}=\frac{3}{6}=\frac{1}{2}=0.5.$$

We often give probability as a percentage – just take the probability we have worked out, multiply it by a hundred and add the per cent sign: 0.5 · 100% = 50%. A probability of fifty per cent tells us that we have the same chance of rolling an even number as of not rolling one. That makes sense, since a dice has three even numbers and three odd numbers.

A certain event has probability 1 and an impossible event has probability 0. Why? We said that for a certain event J we have J = ω, and for an impossible event N we have N = ∅. If we substitute these events into the formula, we get:

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

Worked examples

1. What is the probability of rolling a five on a dice? That is easy: the set of all favourable outcomes is just A = {5}, and the set of all outcomes is still ω = {1, 2, 3, 4, 5, 6}. We substitute into the formula:

   $$P(A)=\frac{|\{5\}|}{|\{1,2,3,4,5,6\}|}=\frac{1}{6}=16.666\dots \mathrm{\%}$$

2. What is the probability of rolling a number divisible by three? We need to find which numbers on a dice are divisible by three. They are only 3 and 6, so the set of favourable outcomes is B = {3, 6}. Then, as usual:

   $$P(B)=\frac{|\{3,6\}|}{|\{1,2,3,4,5,6\}|}=\frac{2}{6}=\frac{1}{3}=33.333\dots \mathrm{\%}$$

3. Now let’s try rolling two dice at the same time. What is the probability of rolling a six on both dice? First we have to work out ω, the set of all outcomes. We have to count all the possibilities that can occur. So if the first dice shows 1, the second can show 1 to 6. That gives six possibilities. If the first dice shows 2, the second dice can again show 1 to 6; that gives another six possibilities. And so on for every number the first dice can show. All the possibilities then look like this:

   $$\begin{array}{rcl}& \omega =& (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),\\ & & (2,1),(2,2),(2,3),(2,4),(2,5),(2,6),\\ & & \dots \\ & & (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}$$

   We have six rows in total, and in each row we have six possibilities. That gives 6 · 6 = 36 possible outcomes for the two dice. How many of them are also in the set of favourable outcomes? Only one, namely (6, 6), because we want two sixes. This gives the probability:

   $$P(C)=\frac{|\{(6,6)\}|}{|\omega |}=\frac{1}{36}=2.777\dots \mathrm{\%}$$

4. What is the probability that two dice show the same number? The number of all possible outcomes is still 36, see the previous example. But how many favourable outcomes are there? Only in six cases do both dice show the same number: D = {(1, 1), (2, 2), …, (6, 6)}. We substitute into the formula:

   $$P(D)=\frac{|D|}{|\omega |}=\frac{6}{36}=\frac{1}{6}=16.666\dots \mathrm{\%}$$

5. What is the probability that, when we roll two dice, one white and one black, the white dice shows 3 and the black dice shows 5? The size of ω is still 36. What does the set of favourable outcomes E look like? It contains just the single outcome (3, 5), so the probability is

   $$P(E)=\frac{|E|}{|\omega |}=\frac{1}{36}=2.777\dots \mathrm{\%}$$

6. What is the probability that, when we roll two dice, one white and one black, they show the numbers 3 and 5? This example is similar to the previous one, except that we do not require the 3 to be on the white dice and the 5 on the black dice. One of the dice just has to show 3 and the other one 5. That gives us more possibilities: besides (3, 5) we also have (5, 3). The probability of this event F is therefore:

   $$P(F)=\frac{|F|}{|\omega |}=\frac{2}{36}=\frac{1}{18}=5.555\dots \mathrm{\%}$$

7. What is the probability that two dice show at least one six? We will go step by step: if the first dice shows a six, the second dice can show 1 to 5. That is five possibilities. If the second dice shows a six, the first dice can show 1 to 5. That is another five possibilities. And finally, both dice can show a six – that is one more possibility. In total we have 5 + 5 + 1 = 11 possibilities. The set of favourable outcomes G therefore 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)}. This gives the probability:

   $$P(G)=\frac{|G|}{|\omega |}=\frac{11}{36}=30.555\dots \mathrm{\%}$$

8. There are 30 students in a class. What is the probability that a student called Oliver is picked in a history lesson, if the teacher always questions two students per lesson and nobody has been picked yet? To work out the probability, we need to calculate how many different pairs of students there are, and how many different pairs have Oliver as one of the two. Since the order in which the teacher picks the students does not matter, we will use combinations.

   Now it will be simple. We have 30 students in total, and we ask how many different pairs we can make from them. That leads us to the binomial coefficient

   $$(\genfrac{}{}{0ex}{}{30}{2})=435$$

   And in how many pairs is Oliver included? If one of the two is Oliver, there are 29 students left that Oliver can be paired with (he cannot be paired with himself, which is why 30-1=29). Now we know the number of all possibilities and the number of favourable possibilities, so we substitute into the formula:

   $$P(H)=\frac{29}{435}=0.0666\dots =6.666\dots \mathrm{\%}$$

   Note: we can also count all the pairs without combinations. How could we form all the pairs from a class of 30 students? By taking one student and adding each of the others in turn. That means we take, for example, Harry and pair him with every other student, which makes 29 pairs. We do this with every student, that is, with all 30 students. This gives 30 · 29 = 870 pairs in total. However, every pair is there twice: once as the pair (Harry, Isla) and once as (Isla, Harry) (depending on whether Harry was the one the rest of the class was paired with, or the one being paired with Isla). So we still have to divide this result by two, and we get the final result: $870÷2=435$.

9. What is the probability that, when you pull three shoes out of a shoe rack holding twelve pairs, all three are left shoes? This is a handy real-life example – I am sure you have needed to know this at some point. We start by working out how many different sets of three shoes we can pull out of the rack. We will need combinations, because the order in which we pull out the shoes does not matter. We have 24 shoes in total and want to pull out 3. This gives the binomial coefficient:

   $$(\genfrac{}{}{0ex}{}{24}{3})=2024$$

   So we have worked out that there are 2,024 different combinations of three shoes that we can pull out of the rack. Now we have to work out which combinations are favourable, that is, which combinations of three shoes contain only left shoes. We know that there are 12 left shoes in the rack altogether. So we just work out how many combinations we can make from these 12 shoes. That again leads to a binomial coefficient:

   $$(\genfrac{}{}{0ex}{}{12}{3})=220$$

   So there are 220 combinations that contain only left shoes. That is everything we need to know, and now we just substitute into the formula:

   $$P(I)=\frac{220}{2024}=0.108695\dots =10.8695\dots \mathrm{\%}$$

   So there is roughly a ten per cent chance of pulling out three shoes that are all left shoes.