More probability

We learned about a number of probability topics in AS Maths including Venn Diagrams, Tree Diagrams and the definitions of Mutual Exclusivity and Independence. In A2 Maths, we take a closer look at conditional probabilities and mutual exclusivity & independence with a particular focus on set notation and introduce some useful formulae.

Conditional Probability

Conditional probabilities are probabilities that change given the occurrence of a previous event or given additional information about an outcome. We have seen conditional probabilities in tree diagrams:

The tree diagram above shows the different outcomes when drawing two counters without replacement from a bag containing 4 red and 6 blue counters – see more Tree Diagram examples. The probabilities change depending on the outcome of the first counter drawn – this is an example of conditional probability. Conditional probabilities may also be presented in a two-way table (or contingency table) – see Example 1. We represent a conditional probability as P(A|B) and we read it as ‘the probability that A occurs given that B has occurred’ or ‘the probability of A given information of B’. A conditional probability question for a tree diagram can also be seen in Example 1.

Set Notation

We can also calculate conditional probabilities using Venn diagrams by using prior information of an outcome to restrict the sample space. Before we do so, we take a brief look at set notation. We saw the following definitions when learning about Venn diagrams in AS Maths. We now introduce the corresponding set notation:


venn diagrams

A^C or A' 


venn diagrams

A\cup B


venn diagrams

A\cap B


venn diagrams

A'\cap B 

Probability Formulae

  1. Addition Formula:

    P(A\cup B)=P(A)+P(B)-P(A\cap B)

    If you colour in event A in one colour, then event B in another, you will notice that the intersection was coloured in twice. Hence, in order to find the probability of the union, one must add the respective probabilities of events A and B whilst removing a single probability corresponding to one of the intersections.
  2. Multiplication Formula:

    P(A|B)=\frac{P(A\cap B)}{P(B)}

    This can be seen in a Venn diagram – given the information that B has occurred, the new full sample space becomes the set B. The probability that A has occurred is then the part of A that lies in the intersection. It can be thought of as the chances of being in A given that you are in B is the probability of the intersection out of the probability of B. It follows from the formula that P(A\cap B)=P(B)\times P(A|B). This can be seen in a tree diagram – given that B has occurred follow event B along the first branch with probability P(B). The probability that A then occurs is P(A|B) following event A along the second branch. Alternatively, P(A\cap B)=P(A)\times P(B|A), which can be seen from the original formula by swapping A and B around.

Mutually Exclusive and Independent Events

We saw the definitions of mutual exclusivity and independence in AS Maths. If events are mutually exclusive, they do not intersect on a Venn diagram. It follows that…

P(A\cap B)=0 for mutually exclusive events.

For independent events, P(A|B)=P(A) – the fact that event B occurred does not affect the probability of A occurring. From the multiplication formation it follows that P(A\cap B)=P(A)\times P(B) and so…

P(A\cap B)=P(A)P(B) for independent events.

See Example 2 for a demonstration of how to use the formulae and how to determine if events are mutually exclusive or independent.


The following two-way table shows the eye and hair colour of 100 university maths students:

Blue EyesBrown EyesTotal
Dark Hair105464
Light Hair142236

Let Bl be the event that a student has blue eyes, Br be the event that a student has brown eyes, D if they have dark hair and L if they have light hair.

a) For a student drawn at random, find the following probabilities:
i) P(Bl)
ii) P(Br)
iii) P(D|Bl)
iv) P(L|Br)

b) Show this information in a tree diagram.

c) Use the tree diagram to find P(Bl\cap L) and P(Bl'\cap L).

i) According to the table, 24 out of 100 students have blue eyes and so P(Bl)=\frac{24}{100}=\frac{6}{25}.
ii) Similarly, P(Br)=\frac{76}{100}=\frac{19}{25}
iii) Of the 24 people that have blue eyes, 10 have dark hair and so the probability of dark hair given blue eyes is \frac{10}{24}, that is, P(D|Bl)=\frac{5}{12}.
iii) Of the 76 people that have brown eyes, 22 have light hair and so the probability of light hair given brown eyes is \frac{22}{76}, that is, P(L|Br)=\frac{11}{38}.

c) By following the branches through the tree diagram, blue eyes then light hair, we see that P(Bl\cap L)=P(Bl)\times P(L|Bl)=\frac{6}{25}\times\frac{7}{12}=\frac{42}{300}=\frac{7}{50}. Similarly, not blue eyes (or brown eyes) then light hair, gives P(Bl'\cap L)=P(Bl')\times P(L|Bl')=P(Br)\times P(L|Br)=\frac{19}{25}\times\frac{11}{38}=\frac{209}{950}=\frac{11}{50}.

Given that P(A)=0.45, P(B)=0.25 and P(A\cup B)=0.65,

a) Find P(A\cap B).
b) Determine whether the events A and B are mutually exclusive and/or independent or not.
c) Find P(A|B).
d) Show the information in a Venn diagram and determine
i) P(A|B') and
ii) P(A\cap B|A\cup B)

a) We use the addition formula: P(A\cup B)=P(A)+P(B)-P(A\cap B) or 0.65=0.45+0.25-P(A\cap B) and rearrange to get P(A\cap B)=0.05.
b) Since P(A\cap B)\ne 0, events A and B are not mutually exclusive. Events are independent if P(A\cap B)=P(A)P(B) – in this case, P(A)P(B)=0.45\times 0.25=0.1125 whereas P(A\cap B)=0.05 and so the events aren’t independent either.
c) Using the multiplication formula: P(A|B)=\frac{P(A\cap B)}{P(B)}, it follows that P(A|B)=\frac{0.05}{0.25}=\frac{1}{5}.

To find P(A|B') we restrict the sample space to not B and look at the part of A that is not in B. This is 0.4 and since it is out of 0.75, we have P(A|B')=\frac{0.4}{0.75}=\frac{8}{15}.

To find P(A\cap B|A\cup B) we restrict the sample space to A\cup B and look at the part of A\cap B that lies in it. All of A\cap B is in A\cup B and so P(A\cap B|A\cup B)=\frac{P((A\cap B)\cap(A\cup B)}{P(A\cup B)}=\frac{P(A\cap B)}{P(A\cup B)}=\frac{0.05}{0.65}=\frac{1}{13}.