Probabilities
Basic Terminology
Contents
This topic introduces the basic concepts required to solve problems involving chance and probability. Within the study of probability there are a number of terms that are important to understand, to be able to interpret questions. We shall now study these words and their meanings.
 Sample Space: The set of all possible outcomes for an experiment.
 Favourable Outcomes: These are those results from an experiment that are required by the question. E.g. in the question: “Find the probability of obtaining a spade when drawing a card from a deck”, the favourable outcome of the experiment is obtaining a spade.
 Mutually Exclusive Outcomes: These are outcomes that cannot occur together. The occurrence of one outcome precludes all others in a set of mutually exclusive outcomes. E.g. consider the gender of a baby, the event of discovering it is a girl precludes it from being a boy. E.g. consider rolling a six sided dice – the events of possible numbers 1,2,3,4,5,6 are all mutually exclusive – rolling a 5 means a 1 was not rolled – again this is a set of mutually exclusive events. Counterexample: rolling a dice twice in succession – if the first roll is a 5, this does not preclude the second roll producing a 5 – thus this is not a set of mutually exclusive events.
 Independent Outcomes: These are outcomes that do not affect each other. E.g. flipping a coin twice in succession produces independent outcomes because flipping a head in the first trial does not affect the probability of a head or tail on the second trial.
 Union: This word is the mathematical form of the word “or”. It is given the symbol .
 Intersection: This word is the mathematical form of the word “and”. It is given the symbol .
Definition of probability
The probability of an event is defined to be the ratio of the number of favourable outcomes to the number of total outcomes, assuming the events are independent. That is, the probability of the event occurring is:
Of course, we assume that every outcome is equally likely to occur. In reality this may not be true for certain (most) events, however this assumption is valid for our purposes. Here we consider the following examples which illustrate this definition.
Example 1Find the probability of obtaining an even number upon rolling a dice. Solution 1Firstly, the favourable outcome is obtaining an even number upon rolling the dice. There are 3 possible even numbers that may be obtained: or . Hence there are favourable outcomes. There are possibilities in total. Hence using the definition of probability we have that:

Example 2Find the probability of obtaining a King from a deck of cards, upon drawing a card. Solution 2There are 4 Kings within the deck of cards, one from each suit. This is the number of favourable outcomes. There are 52 cards in total from which we may choose, making the total number of possible outcomes 52. Hence we have from the definition of probability that:

Example 3In a bag there are red; 3 blue; and white balls. One ball is drawn at random. What is the probability that the ball is: a) Red? b) Red or blue? c) Not red? Solution 3a) As there are 4 red balls and 12 balls in total, then the probability of obtaining a red ball is given by:
b) As there are balls that are either red or blue, and there are 12 balls in total, it follows that the probability of obtaining a red or blue ball is given by:
c) If a ball is not red, then it is either blue or white. Hence as there are blue or white balls then the probability of obtaining a ball that is not red is given by:

Now, since the total number of events that may occur is always greater than or equal to the number of ways a certain event may occur, then it follows that the probability of an event occurring is always less than or equal to one. Also, since a probability is a ratio of two nonnegative numbers, then it follows that the probability of any event satisfies the following inequality:
Now, the last two aspects should be known from studies conducted in former years. The next point is quite important, and is used extensively throughout this course.
The complementary event of an event , say, is denoted by , and is essentially the event where does not occur. E.g. the complementary event of obtaining a king upon drawing a card from a deck of cards, is equal to the event of obtaining a card which is not a king upon drawing a card from a deck of cards. Another example is: the complementary event of obtaining an even number upon rolling a die is equal to obtaining an odd number upon rolling the die.
Now, it should be quite obvious that in any experiment, either an event occurs, or the complementary event occurs (that is to say that either will occur or will not occur). Hence we have that the probability of (read as probability of union complement) is given by . That is:
Now, from the addition rule for probabilities we have that (Not specifically included in course – if students would like more information refer to “A Higher School Certificate Course in Mathematics, Form 6” by Jim Coroneos):
Recall that is the mathematical symbol for “and”. Since we have that (Since both and cannot occur simultaneously) then it follows that:
More formally we have that:
Consider the event . The probability of the event occurring is given bywhere is the complementary event of . 
Note: Some texts use the symbol instead of to represent the complement. That is, instead of writing the complement as , they represent the complement by .
We shall now consider some examples to illustrate use of this rule.
Example 4In a raffle with 1000 tickets, a person buys tickets. Calculate the probability that she does not win first prize. Solution 4To calculate the probability of her to not win is equal to the complement of her winning. Hence we have that:
Hence we have that:

Example 5A card is randomly drawn from a standard deck of cards. Calculate the probability that a heart is not drawn. Solution 5To find the probability that a heart is not drawn we simply calculate the complementary probability of drawing a heart. Since there exist 4 suits with 13 cards each in a standard deck we thus have.
Hence we have that:

Note: The complementary event is usually used when one of “not”, “least” or “most” is used in the question stem. Of course, use of the complementary event is not always possible (or easy) so be sure to check whether the complementary event applies or not.
Mutually Exclusive Events
Mutually exclusive events are events where the occurrence of one precludes all others. To obtain the probability of independent mutually exclusive events occurring sequentially (any occurrence of two events is called a – two stage experiment, in which one event is first, followed by another), one may simply multiply the respective probabilities of each event occurring, to obtain the resultant probability. Mathematically, we have that:
The probability that the independent events and both occur is given by 
This rule is named the multiplication or product rule of probability. Consider the below examples which illustrate use of this.
Example 6What is the probability of rolling a 6 with a dice and randomly drawing a spade from a standard deck cards? Solution 6A standard deck of cards contains 52 cards and 13 spades. Thus the probability of drawing a spade is given by:
Now, the probability of obtaining a 6 with a dice is given by:
Now, these events are independent, and hence we have that the probability of both these events occurring is given by:

Example 7It has been found that development of lung cancer and AIDS have no direct correlation. The probability that someone will develop lung cancer is and the probability that the person will develop AIDS is 0.001. Find the probability that the person in question will contract both AIDS and lung cancer. Solution 7Here we are given that lung cancer and AIDS are independent of each other. Thus we have that:

Example 8In a bag there are 4 green marbles, 5 blue marbles and 3 red marble. John randomly draws a marble, records it and replaces into the bag. Find the probability that: a) The first three marbles drawn are green. b) At least one of the marbles out of the first four drawn is red. Solution 8a) The drawing of marbles in successive experiments is independent in this case as the marbles are being replaced. Hence as we have that the probability of drawing a green marble is it thus follows that the probability of drawing the first three marbles as green is: b) Here we need to consider the complementary event. The complementary event of “At least one of the marbles out of the first four drawn is red” is that “None of the first four marbles drawn are red”. Hence we must firstly find the probability that none of the first four marbles drawn are red. To do this, we simply find the probability that a drawn marble is not red on any single experiment, and then simply raise this number to the fourth power, as we are considering four experiments. That is:
and upon consideration of the complementary event:

Note: The replacement of the marbles in example 7 implies that the probabilities do not change as the stage of the experiment progresses. In general, no replacement implies that the probability of the events is dependent on each other.
Dependent Events
These events are those that are not independent of each other – i.e. the occurrence of one event affects the probability of the occurrence of the other event(s). For example, drawing coloured balls randomly out of a bag without replacement. There are various ways to analyse such events; the most common being simple use of counting techniques which take into consideration the changes that the previous event has brought about. We shall now illustrate examples of dependent events, and methods which we can use to solve problems involving them.
Example 9A bag contains red and green discs. One disc is drawn out and then another disc is drawn. What is the probability that: a) The discs are the same colour?
b) The first disc is red and the second is green? Solution 9a) To find the probability that the two discs of the same colour are drawn, we firstly need to find the probability either both discs are green or both discs are red. Now, consider the case where both of the discs are green. In this case we have that after initially picking a green disc, of which there are ten, there are then nine green discs left with a total of fourteen left in the bag. Hence the probability of obtaining a green disc on the second pick then changes to . Now, we may obtain the resultant probability by multiplying the respective probabilities found.
Similarly for the red discs, initially there are 5 red discs from 15 in total. After a red disc is picked, there are 4 remaining, from a total number of 14. Hence we have that:
Now, to obtain the probability of obtaining discs of the same colour, we simply add the probabilities found above. Doing so gives:
b) To find the probability that the first disc is red and that the second disc is green we simply consider the probabilities before and after the first event occurs as above. Now, the probability of choosing a red disc first is equal to as the number of discs at that point is equal to that stated in the question. Now, the probability of then choosing a green is equal to as there then remain discs with a total of green remaining. Hence we may obtain the probability required by multiplying the results:

Note: Students often express confusion in knowing when to add probabilities and when to multiply probabilities (As in the above example). A general way to solve this dilemma is to consider the conjunction, combining the two events in the question (in the case of a twostage experiment). That is, if the conjunction combining the two events is an “or” (Consider example 7 part (a) which combines the events “both green” and “both red” with the conjunction “or”) then the probabilities are added, otherwise (i.e. if the conjunction is “and”) then the probabilities are multiplied (Consider example 7 part (b) which combines the events with the conjunction “and”). Be careful though, probabilities cannot be added if there is an overlap between the two experiments. Consider example 10 part (d) to illustrate the overlap.
Example 10A bag contains twelve jelly beans. Three are coloured red, four black and five green. Peter eats three jelly beans chosen randomly from the bag. Find the probability that: a) The first jelly bean is black; b) All three jelly beans are black; c) Exactly one of the jelly beans is black. Solution 10a) The probability that the first jelly bean is black is simply the probability of pulling a black jelly bean out upon randomly choosing a jelly bean from the bag. i.e. we have that:
b) The probability that all three jelly beans are black is simply given by:
Since upon removing a black jelly bean, the number of black jelly beans remaining is decreased, as is the number of total jelly beans.
c) To find the probability that exactly one of the jelly beans is black involves three parts. The black jelly bean may occur as any one of the first, second or third jelly bean.
Hence we have that the probability of obtaining exactly one black jelly bean is:

Note: For those students studying Mathematics Extension 1, you may have used combinations to great effect in the last example to count the total number of ways that the black jelly may have occurred once.
Example 11A card is drawn from a deck of cards. What is the probability that: a) A red card is drawn? b) An even numbered card (2 , 4 , 6 , 8 , 10) is drawn? c) A red, even numbered card is drawn? d) A red or an even numbered card is drawn? Solution 11a) Since there are red cards in a pack of cards, we have that:
b) Since there are even numbered cards in each suit, and there exists suits in total, then there are even cards in total. Hence we have that the probability of obtaining an even cards is given by:
c) There are two red suits (hearts and diamonds) and within each of these there exists even cards. Hence there are red even cards in total. Thus we have:
d) There are in total, red cards, of which are even cards. Now, there are also even cards that are not red. Hence in total there are cards that are either red, even or both.
Note here that we cannot simply add the probabilities obtained in (b) and (c), since there is an overlap in the defining conditions for the favourable outcomes. i.e. there exists red cards that are even. 
Tree Diagrams
In the case where there are many outcomes and many experiments linked together, it may be useful to represent the information diagrammatically. One such way to represent the probabilities diagrammatically is through a tree diagram, called so because the outcomes branch out from the original result like the branches of a tree.
To construct a tree diagram for a multistage experiment, simply determine which experiments are happening and in which order, after which the possible outcomes of the experiment are listed. From here, branch out the possible outcomes of the next experiment from each outcome in the previous experiment, and simply continue to do so for the next experiment and so on. Now, as a convention, write the probabilities of each event occurring on the branch itself. To obtain the probability of an event, simply find the branch along which the event occurs, and then find the probability by multiplying together the probabilities along that continuous branch. If there is more than one possible branch that is the required outcome, simply find the probability of this outcome occurring by multiplying along the branches, and then add this result to all the other possible branch probabilities that are favourable.
Consider the following examples to illustrate the techniques required.
Example 12A bag contains red marbles and blue marbles. A marble is drawn out, its colour noted, and then replaced. Another marble is then drawn out. Calculate the probability of selecting different coloured marbles. Solution 12We firstly draw the required tree diagram. The branches required are shown with weighted lines. Now to obtain the required probabilities, we simply multiply the probabilities along the branches heading from the branch point. As we are to add probabilities occurring on separate branches, we have that:

Consider the next two examples, illustrating the use of tree diagrams for dependent experiments.
Example 14A bag contains 3 green marbles and 7 blue marbles. What is the probability of selecting marbles of different colours if the first marble is not replaced before drawing the second? Solution 14Firstly we draw up the tree diagram after considering the alterations in the probabilities after removing certain marbles (Remember that this is a twostage dependent experiment). Now, the cases we require (different colours) are outlined with broader lines. Hence we have that the probability of obtaining different colours is:

It can be seen that both independent and dependent events may be represented using tree diagrams.
Lattice Diagrams
In this last section, we shall look at another diagrammatic representation of probabilities. In particular, we shall be considering two stage experiments in which the experiments are independent, which may be represented using a table.
Lattice diagrams are useful for finding probabilities of certain independent events occurring that otherwise involve very tedious calculation. E.g. the question “A person rolls two dice. Find the probability that the numbers obtained have a sum of 6,” may be easily solved using the techniques taught in this section.
To solve such problems, we simply list the possible outcomes of the first experiment vertically and then we list the possible outcomes of the second experiment horizontally. This then gives us a table format, and within each grid there exists correspondence to a certain outcome from the first experiment and a certain outcome from the second. We then simply combine the two events and write the required form of the final outcome of the two stage experiment, and this becomes one of the possible outcomes. Consider the following example to illustrate the notions involved.
Example 15:Find the probability of obtaining a sum of 7 upon rolling two dice.Solution 15:So, we firstly list the outcomes of the two separate events (1 , 2 , 3 , 4 , 5 , 6) separately, both vertically and horizontally.
We now, consider the total outcome corresponding to every possible “coordinate” or pair of events, in this case the outcome required is the sum of the numbers. We also now represent the diagram in tabulated form.
Now it can be seen that the sum of occurs times, out of the total sums that may occur. Hence we have that the probability of obtaining a sum of upon rolling two dice is given by:

Note: This method may be used for all twostage experiments involving independent events however, it must not be used for events involving dependent events. In this case, the best method is to use a tree diagram.