# Probability an Introduction

See also: Estimation, Approximation and RoundingProbability is the science of how likely events are to happen. At its simplest, it’s concerned with the roll of a dice, or the fall of the cards in a game. But probability is also vital to science and life more generally.

Probability is used, for example, in such diverse areas as weather forecasting and to work out the cost of your insurance premiums.

A basic understanding of probability is an essential skill in life, even if you are not a professional gambler or weather forecaster.

## Basic Probability: Some Concepts

The probability that an event will occur is a number between 0 and 1. In other words, it is a fraction. It is also sometimes written as a percentage, because a percentage is simply a fraction with a denominator of 100. For more about these concepts, see our pages on Fractions and Percentages.

**An event that is certain to occur has a probability of 1, or 100%, and one that will definitely not occur has a probability of zero. It is also said to be impossible.**

What is Probability?

The probability (P) that an event will happen is:

Probability is easier to understand with an example:

Suppose that you are going to throw a standard dice, and you want to know what your chances are of throwing a 6.

In this case, there is only **one outcome** that leads to that event (ie. you throw a 6), and 6 possible outcomes altogether (you might throw 1, 2, 3, 4, 5 or 6).

**The probability of throwing a six is therefore ^{1}/_{6}.**

Now suppose that you want to know what your chances are of throwing 1 or 6. Now there are **two favourable outcomes**, 1 and 6, but still 6 possible outcomes.

The probability is therefore ^{2}/_{6}. Which you can reduce down to ^{1}/_{3}.

For more about reducing fractions, see our page on **Fractions**.

### Probability of Multiple Events

Probability gets a bit more complicated when you have multiple events, for example, when you’re tossing more than one coin, or throwing several dice.

The reason is that you have more possible outcomes.

For example, when you are tossing two coins, each one could land heads or tails up. So instead of just two possible outcomes (heads or tails), there are now four:

First Coin | Head | Head | Tail | Tail |

Second Coin | Tail | Head | Tail | Head |

More coins will mean more possible outcomes.

As a rule of thumb, the number of possible outcomes is equal to:

The number of outcomes per item to the power of the number of items.

So if you have five coins, each with two possible outcomes, the total number of possible outcomes is 2^{5} = 2 x 2 x 2 x 2 x 2 = 32.

If you want to work out the probability of throwing a head and a tail when you throw two coins, there are two outcomes that are favourable (the first coin is heads, and the second is tails, or the first is tails and the second is heads), and four events in total. The probability is ^{2}/_{4} or ^{1}/_{2}.

Top Tip!

Most mistakes in probability lie in either not calculating the true number of possible outcomes, or not calculating the true number of favourable outcomes.

Always take time to ensure that you have all the possible outcomes. If necessary, list them.

Worked example

If you throw three dice, what is the probability that you do not throw any 4s, 5s, or 6s?

You are throwing three dice, each of which has 6 possible outcomes.

The total number of outcomes is therefore 6^{3} = 6 x 6 x 6 = **216**

Each dice has **three** favourable outcomes, 1, 2, or 3.

For the first two dice, you need to throw either 1, 2, or 3 for both dice. The favourable outcomes are:

*1-1 1-2 1-3 2-1 2-2 2-3 3-1 3-2 3-3*

In other words there are **nine** favourable outcomes with two dice. Now each one of these has three possible favourable outcomes from the third dice (ie. the third dice could be 1, 2, or 3).

So the number of favourable outcomes is 9 x 3 = **27**.

The probability of not rolling 4, 5, or 6 with three dice is therefore ^{27}/_{216} = ^{1}/_{8}.

## Independent and Dependent Probability

The above rules apply when the items are **independent**, for example, dice or coins, and the outcome of the first one does not affect the second or subsequent events.

However, it gets more complicated when the first event affects the second and subsequent events, that is, they are **dependent**.

## Dependent Probability

Probability of multiple events when the first event affects the second.

**Dependent events** are not as unusual as you might think. Consider drawing cards from a pack. If you do not replace the cards after each draw, you have a different number of possible outcomes each time. In this case, you need to work out the probability of each event happening and then combine them in some way.

The way that you combine them depends on whether you want to know the probability of **either** event, or **both** events (** OR** or

**):**

*AND*- To work out the probability of
**both**events*(AND)*, you**multiply**the probability of one by the probability of the other. - To work out the probability of
**either**event*(OR)*, you**add**the probability of one to the probability of the other.

Worked example

What is the probability of drawing at least one ace from a pack of cards on two draws, if you do not replace the cards in between?

There are 52 cards in the pack, four of which are aces.

**There are three possible favourable outcomes:**

You could draw two aces - Ace/Ace

Or you could draw one ace, either as the first or second card - Ace/Not, Not/Ace.

In AND/OR terms, these are:

*Ace AND Ace OR**Ace AND Not Ace OR**Not Ace AND Ace.*

This means that to solve the problem we need to use both multiplication and addition.

**The first scenario: Ace and Ace**

The probability of drawing an ace on the first card is ^{4}/_{52} = ^{1}/_{13}.

Once you have drawn one ace, there are only 51 cards left from which to draw the second card, and only three of them are aces. The probability of drawing a second ace is therefore 3/51. You want both events, so you need to multiply them.

The probability of drawing Ace AND Ace is ^{1}/_{13} x ^{3}/_{51} = ^{1}/_{221}

**The second scenario: Ace and Not Ace**

The probability of drawing an ace remains ^{1}/_{13}. But now you have 51 cards left, all but three of which are not aces. 51−3=48.

Your chance of drawing a ‘not ace’ on the second card is therefore ^{48}/_{51}, and the chance of drawing Ace AND Not Ace is ^{1}/_{13} x ^{48}/_{51} = ^{16}/_{221}

**The third scenario: Not Ace and Ace**

The probability of drawing a ‘not ace’ on the first card is (52-4)÷52 = ^{48}/_{52}

The probability of drawing an ace on the second card is ^{4}/_{51}.

The probability of drawing Not Ace AND Ace is therefore ^{48}/_{52} x ^{4}/_{51} = ^{16}/_{221}

*Note that in this case, this is the same as Ace-Not Ace. *

*This will not always follow for all scenarios.*

The overall probability

The probability of drawing at least one ace when you draw two cards is therefore the probability of each of the three scenarios added together (because you need only one to happen: they are OR events).

**The answer is ^{1}/_{221} + ^{16}/_{221} + ^{16}/_{221} = ^{33}/_{221}.**

Top Tip!

If you have trouble remembering whether you have to add or multiply for AND or OR, here are two easy ways to remember:

For AND, you don’t add.

The probability of throwing either a head or a tail from one coin is 1 (it is a certainty). The probability of each outcome is ½. If you multiplied these, you would get ¼. You don’t. You add them: ½ + ½ = 1.

It’s also worth remembering that the total probability cannot be more than 1. If your answer is greater than 1, you have probably added instead of multiplying.

### A Word of Reassurance

Advanced probability sums can get extremely lengthy by the time you’ve written out all the possible outcomes. However they are no more difficult to do. As long as you correctly work out all the favourable outcomes, and all the possible outcomes, all you have to do is plug the numbers into the AND/OR formula and you will get the right answer.

Continue to:

Simple Statistical Analysis

Set Theory | Simple Equations