**Introduction to Probability**

We use capital letters to represent **events**, e.g. X = score of 5 on 2 dice, or Y = rain on Wednesday.

P(X) means the probability that X will happen, e.g. P(X)=1/9.

Events are often made up of outcomes, e.g. event X is made up of the outcomes {(1,4),(2,3),(3,2),(4,1)}.

For any event, 0 ≤ P(X) ≤1.

If a selection is **random**, it means that all **outcomes** are equally likely.

This means that if we select 1 item from n items, then each outcome has a probability of 1/n of being selected.

The basic formula for the probability of an event, with equally likely outcomes is:

**Complementary Probability**

If X is the kind of event that either happens or doesn’t happen (e.g. a coin landing on Heads), then: P(A) + P(A’) = 1.

In this case we say that A and A’ are **complementary** events.

We may write probabilities either as simplified fractions or as decimals rounded to 3 significant figures.

**Probability Introduction Example**

A circular spinner is divided into 5 sections with angles of 36º, 54º, 72º, 90º and 108º. The sectors are labelled respectively with the numbers 2, 4, 6, 7, 8.

- What is the probability that and even number is obtained with each spin?
- How many times should we expect an odd number to be obtained with 60 spins?

**Exercise**

**Answers**

**Mutually Exclusive Events**

Events X and Y are mutually exclusive if they cannot both occur, so P(X and Y)=0 (e.g. X = person is a man, Y = person is a woman).

If events are mutually exclusive, the probability of one of the events happening or the other event happening is equal to the sum of the events’ probabilities.

e.g. P(a playing card is a ‘3’ or a ‘5’) = P(card is a ‘3’) + P(card is a ‘5’).

Note that the word **or** in mathematics is inclusive, unlike in typical English (i.e. if you say you will invite Susan or Samantha to the cinema, it is not inclusive).

By contrast, if events are not mutually exclusive, the probability of one of the events happening or the other event happening **is not** equal to the sum of the events’ probabilities.

e.g. P(a playing card is a ‘3’ or a ‘red’) ≠ P(card is a ‘3’) + P(card is a ‘5’).

In general, **for mutually exclusive events**, P(X or Y) = P(X) + P(Y), and similarly, P(X or Y or Z or …) = P(X) + P(Y) + P(Z) + …. For events that **are not mutually exclusive**, P(X or Y) = P(X) + P(Y) – P(X and Y).

We can use set notation ∪ instead of ‘or’ and ∩ instead of ‘and’.

Venn diagrams can be a useful tool for solving problems:

**Worked Example**

A fair 12 sided die with faces numbered from 1 to 12 is rolled. Find the probability that an even number or a factor of 20 is obtained.

**Worked Example 2**

In a survey, 50% of the participants own a desktop (*D*), 60% own a laptop (*L*) and 15% own both.

What percentage of the participants owns neither a desktop nor a laptop?

**Exercise**

**Answers**

**Independence**

Events X and Y are **independent** if the probability of event X happening isn’t affected by event Y happening.

For independent events, P(X and Y) = P(X) x P(Y), or in general, P(X and Y and Z and …) = P(X) x P(Y) x P(X) x …

A common example of independent events are items that are selected **with replacement**.

Tree diagrams are a very useful way to calculate and display probabilities where there are multiple **trials**. e.g:

**Possibility Diagram / Sample Space**

If there are only two trials, a possibility diagram can also be very useful for calculating probabilities. e.g.:

**Worked Example**

In a game, a participant has to spin a fair coin and then roll a fair dice with faces numbered from 1 to 6. If the coin shows heads and the dice score is below 3, the participant wins a prize. What is the probability that they win a prize?

**Worked Example 2**

You select 2 cards from an ordinary pack of cards. What is the probability that they are both not kings?

**Worked Example 3**

**Exercise**

**Answers**

**Testing for Independence**

P(X and Y) = P(X) x P(Y) <=> Events X and Y are independent.

**Worked Example**

A group of 80 teenagers were each asked if they owned a skateboard. The data was compiled in the following table:

Boys (B) | Girls (B’) | Totals | |

Owns skateboard (S) | 12 | 8 | 20 |

Doesn’t own skateboard (S’) | 36 | 24 | 60 |

Total | 48 | 32 | 80 |

One teenager is selected at random. Show that the event B and the event S are independent.

**Exercise**

**Answers**

**Conditional Probability**

P(X|Y) is read as P (X given Y) and is the probability that event X will occur, given that we already know that event B occurs.

X = number is even, Y=number is multiple of 4

P(X|Y) = 1

P(Y|X) = 1/2

**Worked Example** **1**

A child is selected at random from a group of 11 boys and nine girls, and one of the girls is called Davina. Find the probability that Davina is selected, given that a girl is selected.

**Worked Example** **2**

A group of 45 students from three different schools, X, Y and Z, were asked how many subjects they are studying at A level. The results are given in the following table:

School X | School Y | School Z | Totals | ||

Number of A levels: | 2 | 6 | 5 | 7 | 18 |

Number of A levels: | 3 | 7 | 5 | 8 | 20 |

Number of A levels: | 4 | 4 | 1 | 2 | 7 |

TOTALS | 17 | 11 | 17 | 45 |

One of the students is selected at random. Find the probability that this student:

- studies four A levels, given that the student is from school Z;
- is from school X, given that the student is studying fewer than four A levels.

**Exercise**

**Solutions**

**Independence and Conditional Probability**

If events X and Y are independent, then P(X|Y) = P(X|Y’) = P(X), i.e. the probability of event X occurring is the same whether event Y occurs or not.

Note that events that are not mutually exclusive may still be independent.

Tree Diagrams can be useful calculation aids when working with conditional probabilities.

**Theory**:

So: P(A|B) x P(B) = P(B|A) x P(A)

**Worked example 1**

C and D are two events such that:

- P(C) = 0.2;
- P(D) = 0.6; and
- P(C|D) = 0.3.

Find P(D|C).

**Worked Example 2**

**Worked Example 3**

An ordinary fair die is rolled. Find the probability that the number obtained is prime, given that it is odd.

**Worked Example 4**

A boy walks to school (W.) 60% of the time and cycles (C) 40% of the time. He is late to school (L), on 5% of the occasions that he walks, and he is late on 2% of the occasions that he cycles.

Given that he is late to school, find the probability that he cycles; that is, find P(*C *| *L*).

**Exercise**

**Mixed Exercise**

**Answers to Exercise**

**Answers to Mixed Exercise**