Probability

UK Secondary (7-11)

Mathematical Notations for Probability

Lesson

The notations commonly used in probability are related to *function *notation and also to the notations of *set theory*. This is because the range of possible outcomes of an experiment or observation is a *set*. We assign numbers - called probabilities - to subsets of these outcomes in the manner of a function.

Thus, the notation $P(A)$`P`(`A`) means 'The probability that outcome $A$`A` occurs'. Expressed differently, it means that a unique number is to be assigned to outcome $A$`A`, subject to the constraints of the particular situation.

The historical record on the mathematical treatment of probability includes the 17th-century mathematicians Blaise Pascal and Pierre de Fermat. In the beginning, the subject was motivated by questions about games of chance and it was not until the 20th century that probability as an abstract mathematical idea was put on a firm foundation. Russian mathematician Andrei Kolmogorov. achieved this by devising three axioms of probability from which the familiar ideas about probability can be deduced. For the axioms, some notations of set theory are used, which we now introduce.

The set of all possible outcomes of an experiment or observation is called the *sample space* or *outcome space *and it is often given the symbol $\Omega$Ω. An *event *can be thought of as any subset of the space of outcomes under consideration. Thus, in a die-tossing experiment, the sample space might be the set $\Omega=\left\{1,2,3,4,5,6\right\}$Ω={1,2,3,4,5,6}. The event 'An odd number was thrown' is then represented by the subset $\left\{1,3,5\right\}${1,3,5}.

The symbol $\cup$∪ is used in place of the word 'or'. It is the symbol for set union. The union of two sets is the collection of elements that belong to either set or both.

The symbol $\cap$∩ is used in place of the word 'and'. It is the symbol for the intersection of sets. The intersection of two sets is the subset of elements that belong to both of them.

Kolmogorov's first axiom says: The probability of an event is a positive real number. In symbols this is, $P(E)\in R$`P`(`E`)∈`R`, $P(E)\ge0$`P`(`E`)≥0 where $E\subset\Omega$`E`⊂Ω.

The second axiom says: $P\left(\Omega\right)=1$`P`(Ω)=1 meaning the probabilities assigned to the events belonging to a sample space are not greater than $1$1.

The third axiom relates the ideas of disjoint events and the addition of probabilities. Two events are disjoint or mutually exclusive if they have no outcomes in common. In symbols, the fact that events $A$`A` and $B$`B` are disjoint is expressed by $A\cap B=\varnothing$`A`∩`B`=∅ where the symbol $\varnothing$∅ for the empty set means the absence of an outcome. In the die-tossing example, if events $A$`A` and $B$`B` are respectively the occurrence of odd and even numbers, the events are disjoint because there can occur no number that is both odd and even - the sets of odd and even numbers do not overlap. We say the intersection of $A$`A` and $B$`B` is empty.

The third axiom says: Given any finite or countably infinite sequence of disjoint events $E_1,E_2,E_3,...$`E`1,`E`2,`E`3,... in the sample space, the probability of $E_1$`E`1 or $E_2$`E`2 or $E_3$`E`3 or $...$... is $P(E_1)+P(E_2)+P(E_3)+...$`P`(`E`1)+`P`(`E`2)+`P`(`E`3)+...

That is, $P\left(E_1\cup E_2\cup E_3\cup...\right)=P(E_1)+P(E_2)+P(E_3)+...$`P`(`E`1∪`E`2∪`E`3∪...)=`P`(`E`1)+`P`(`E`2)+`P`(`E`3)+...

It is useful to have a notation for the probability that an event does *not* occur. For an event $E$`E`, we indicate the non-occurrence of $E$`E` by $E'$`E`′ or by $E^c$`E``c` where the superscript $c$`c` stands for 'complement'. Sets $E$`E` and $E'$`E`′ are *complementary *in the sense that $E\cap E'=\varnothing$`E`∩`E`′=∅ and $E\cup E'=\Omega$`E`∪`E`′=Ω.

We will often need a notation for what is called *conditional* probability. To illustrate the idea, suppose a number is to be randomly selected from the space $\Omega=\left\{1,2,3,...,100\right\}$Ω={1,2,3,...,100}. Event $A$`A` is defined to be, 'a number in the range 20 to 50 is chosen'. Event $B$`B` is the event that the number is a multiple of 3. We might ask: What is the probability of $B$`B` occurring given that $A$`A` has occurred? This conditional probability is notated $P(B|A)$`P`(`B`|`A`). In effect, the sample space has been reduced from the original $\Omega$Ω to the set $\left\{20,21,22,...,50\right\}${20,21,22,...,50}.

Show from the axioms that $P(E')=1-P(E)$`P`(`E`′)=1−`P`(`E`).

Since $E\cap E'=\varnothing$`E`∩`E`′=∅, we have from the third axiom, $P(E\cup E')=P(E)+P(E')$`P`(`E`∪`E`′)=`P`(`E`)+`P`(`E`′). That is, $P(\Omega)=P(E)+P(E')$`P`(Ω)=`P`(`E`)+`P`(`E`′). But, $P(\Omega)=1$`P`(Ω)=1. So, $P(E)+P(E')=1$`P`(`E`)+`P`(`E`′)=1 and hence, $P(E')=1-P(E).$`P`(`E`′)=1−`P`(`E`).

In a sample space $\Omega$Ω, $A$`A` and $B$`B` are two events. Suppose that we know $P(A\cap B)=u$`P`(`A`∩`B`)=`u` and we are given that $P(B)=v$`P`(`B`)=`v`. Give an argument explaining what the value of $P(A|B)$`P`(`A`|`B`) should be.

The probability assigned to $A\cap B$`A`∩`B` in the whole space is $u$`u`, out of the total probability $1$1. We need instead to consider the probability of $A\cap B$`A`∩`B` in the smaller space $B$`B` whose probability is $v$`v` in the whole space. So, instead of the proportion $\frac{u}{1}$`u`1, we should have $\frac{u}{v}$`u``v`. Thus, $P(A|B)=\frac{P(A\cap B)}{P(B)}$`P`(`A`|`B`)=`P`(`A`∩`B`)`P`(`B`).

Two events are defined as:

Event $A$`A`: it will rain tomorrow

Event $B$`B`: there will be a storm tomorrow

The notation $P(A\cup B)$`P`(`A`∪`B`) is suitable to describe which of the following probabilities?

probability of no rain tomorrow

Aprobability of there being either a storm or rain tomorrow

Bprobability of a storm occurring tomorrow

Cprobability of there being a storm tomorrow but no rain

Dprobability of no rain tomorrow

Aprobability of there being either a storm or rain tomorrow

Bprobability of a storm occurring tomorrow

Cprobability of there being a storm tomorrow but no rain

D

How could the probability of the event $A$`A`: "getting a number greater than 3 when a die is rolled" be written using probability notation? Select all the correct options.

$P\left(A\right)$

`P`(`A`)A$P$

`P`$($(rolling a die$)$)B$50$50

C$1-P\left(A'\right)$1−

`P`(`A`′)D$P$

`P`$($(greater than 3$)$)E$E\left(A\right)$

`E`(`A`)F$P\left(A\right)$

`P`(`A`)A$P$

`P`$($(rolling a die$)$)B$50$50

C$1-P\left(A'\right)$1−

`P`(`A`′)D$P$

`P`$($(greater than 3$)$)E$E\left(A\right)$

`E`(`A`)F

Consider the two events:

A: Paul wins the golf tournament

B: Paul wins the badminton tournament

The probability that Paul wins either the golf or badminton but not both can be represented by:

$P\left(A'\cap B'\right)$

`P`(`A`′∩`B`′)A$P\left(\left(A\cap B'\right)\cup\left(B\cap A'\right)\right)$

`P`((`A`∩`B`′)∪(`B`∩`A`′))B$P\left(A\cup B\right)$

`P`(`A`∪`B`)C$P\left(A'\cap B\right)$

`P`(`A`′∩`B`)D$P\left(A'\cap B'\right)$

`P`(`A`′∩`B`′)A$P\left(\left(A\cap B'\right)\cup\left(B\cap A'\right)\right)$

`P`((`A`∩`B`′)∪(`B`∩`A`′))B$P\left(A\cup B\right)$

`P`(`A`∪`B`)C$P\left(A'\cap B\right)$

`P`(`A`′∩`B`)D