# How to Find the Probability of an Event and Calculate Odds, Permutations and Combinations

## What Is Probability Theory?

Probability theory is an interesting area of statistics concerned with the odds or chances of an event happening in a trial, e.g. getting a six when a dice is thrown or drawing an ace of hearts from a pack of cards. To work out odds, we also need to have an understanding of permutations and combinations. The math isn't terribly complicated, so read on and you might be enlightened!

What's covered in this guide:

- Equations for working out permutations and combinations
- Expectation of an event
- Addition and multiplication laws of probability
- General binomial distribution
- Working out the probability of winning a lottery

## Definitions

Before we get started let's review a few key terms.

**Probability**is a measure of the likelihood of an event occurring.- A
**trial**is an experiment or test. E.g., throwing a dice or a coin. - The
**outcome**is the result of a trial. E.g., the number when a dice is thrown, or the card pulled from a shuffled pack. - An
**event**is an outcome of interest. E.g., getting a 6 in a dice throw or drawing an ace.

## What Is the Probability of an Event?

There are two types of probability, empirical and classical.

If A is the event of interest, then the probability of A occurring is denoted by P(A).

### Empirical Probability

This is determined by carrying out a series of trials. So, for instance, a batch of products is tested and the number of faulty items is noted plus the number of acceptable items.

If there are n trials

and A is the event of interest

Then if event A occurs x times

P(A) = x / n

**Example:** A sample of 200 products is tested and 4 faulty items are found. What is the probability of a product being faulty?

So x = 4 and n = 200

Therefore P(faulty item) = 4 / 200 = 0.02

### Classical Probability

** **This is a theoretical probability which can be worked out mathematically.

If A is the event, then

P(A) = Number of ways the event can occur / The total number of possible outcomes

**Example 1:** What are the chances of getting a 6 when a dice is thrown?

In this example, there is only 1 way a 6 can occur and there are 6 possible outcomes, i.e. 1, 2, 3, 4, 5 or 6.

So P(6) = 1/6

**Example 2:** What is the probability of drawing a 4 from a pack of cards in one trial?

There are 4 ways a 4 can occur, i.e. 4 of hearts, 4 of spades, 4 of diamonds or 4 of clubs.

Since there are 52 cards, there are 52 possible outcomes in 1 trial.

So P(4) = 4 / 52 = 1 / 13

## What Is the Expectation of an Event?

Once a probability has been worked out, it's possible to get an estimate of how many events will likely happen in future trials. This is known as the expectation and is denoted by E.

If the event is A and the probability of A occurring is P(A), then for N trials, the expectation is:

E = P(A) N

For the simple example of a dice throw, the probability of getting a six is 1/6.

So in 60 trials, the expectation or number of expected 6's is:

E = 1/6 x 60 = 10

Remember, the expectation is not what will actually happen, but what is likely to happen. In 2 throws of a dice, the expectation of getting __a__ 6 (not two sixes) is:

E = 1/6 x 2 = 1/3

However, as we all know, it's quite possible to get 2 sixes in a row, even though the probability is only 1 in 36 (see how this is worked out later). As N becomes larger, the actual number of events which happen will get closer to the expectation. So for example when flipping a coin, if the coin isn't biased, the number of heads will be closely equal to the number of tails.

## Success or Failure?

The probability of an event can range from 0 to 1.

Remember

P(Event) = Number of ways the event can occur / The total number of possible outcomes

So for a dice throw

P(getting a number between 1 and 6) = 6 / 6 = 1 (since there are 6 ways you can get "a" number and 6 possible outcomes)

P(getting a 7) = 0 / 6 = 0 (there are no ways the event 7 can occur in any of the 6 possible outcomes)

P(getting a 5) = 1 / 6 (only 1 way of getting a 5)

If there are 999 failures in 100 samples

Empirical probability of failure = P(failure) = 999/1000 = 0.999

A probability of 0 means that an event will never happen.

A probability of 1 means that an event will definitely happen.

In a trial, if event A is a success, then failure is not A (not a success)

And P(A) + P(not A) = 1

## Addition Law of Probability

For mutually exclusive (they can't occur simultaneously) events A and B

P(A or B) = P(A) + P(B)

**Example:** A dice is thrown and a card is drawn from a pack, what is the possibility of getting a 6 *or* an ace?

P(getting a six) is 1/6

There are 52 cards in a pack and four ways of getting an ace.

P(getting an ace) is 4/52 = 1/13

So P(getting a six

orgetting an ace) = P(getting a six) + P(getting an ace) = 1/6 + 1/13

Remember in these type of problems, how the question is phrased is important. So the question was to determine the probability of one event occurring "*or*" the other event occurring and so the addition law of probability is used.

## Multiplication Law of Probability

For independent (the first trial doesn't affect the second trial) events A and B

P(A and B) = P(A) x P(B)

**Example: ** A dice is thrown and a card drawn from a pack, what is the probability of getting a 5 *and* a spade card?

P(getting a 5) = 1/6

There are 52 cards in the pack and 4 suits or groups of cards, aces, spades, clubs and diamonds. Each suit has 13 cards, so there are 13 ways of getting a spade.

So P(drawing a spade) = 13/52 = 1/4

P(getting a 5

anddrawing a spade) = P(getting a 5) x P(drawing a spade) = 1/6 x 1/4 = 1/24

Again it's important to note that the word "*and*" was used in the question, so the multiplication law was used.

### Independent and Dependent Events

Events are independent when the occurrence of one event doesn't affect the probability of the other event.

So if a card is drawn from a pack, the probability of an ace is 4/52 = 1/13.

If the card is replaced, the probability of drawing an ace is still 1/13.

Events are dependent if the occurrence of the first event affects the probability of occurrence of the second event.

If an ace is drawn from a pack and not replaced, there are only 3 aces left and 51 cards remaining, so the probability of drawing a second ace is 3/51.

For two events A and B where B depends on A, the probability of Event B occurring after A is denoted by P(B|A).

## Permutations and Combinations

To solve more difficult problems and derive an expression for the probability of a general binomial distribution, we need to understand the concept of permutations and combinations. I won't go into the mathematics of the derivation, but basically the expression is derived from the equation for working out combinations.

## A Permutation Is an Arrangement

A permutation is a way of *arranging* a number of objects. So, for instance, if you have the letters ** A, B, and C** then all the possible permutations are:

ABC, ACB, BAC, BCA, CAB, CBA

Note that BA is a different permutation to AB.

If you have n objects, there are *n factorial* number of ways of arranging them, written as n!

n! = n x (n-1) x (n-2) .... x 3 x 2 x 1

The reason for this is because for the first position, there are n choices, and for each of these choices, there are (n-1) choices for the second place (because 1 choice was used up for the first place), and for each of the choices in the first two places, (n-3) choices for the third place and so on.

In the example above, the 3 letters A, B, C could be arranged in 3! = 3 x 2 x 1 = 6 ways

In general, if n objects are selected r at a time then, the number of permutations is:

n! / (n-r)!

This is written as ^{n}P_{r}

**Example:** 2 letters are chosen from the set of letters A, B, C, D. How many ways can the 2 letters be arranged?

There are 4 letters so n =4 and r = 2

^{n}P_{r}=^{4}P_{2 }= 4! / (4 - 2)! = 4! / 2! = 4 x 3 x 2 x 1 / 2 x 1 = 12

## A Combination Is a Selection

A combination is a way of selecting objects from a set without regard to the order of the objects. So again if we have the letters A, B and C and select 3 letters from this set, there is only 1 way of doing this, i.e. select ABC.

If we select 2 letters at a time from ABC, all the possible selections are:

AB, AC, and BC

Remember, BA is the same selection as AB etc.

In general, if you have n objects in a set and make selections r at a time, the total possible number of selections is:

^{n}C_{r}= n! / ((n - r)! r!)

**Example: ** 2 letters are chosen from the set ABCD. How many combinations are possible?

There are 4 letters so n = 4 and r = 2

^{n}C_{r}=^{4}C_{2 }= 4! / ( (4 - 2)! x 2!) = 4! / (2! x 2!)= 4 x 3 x 2 x 1 / ( (2 x 1) x (2 x 1) ) = 6

## General Binomial Distribution

In a trial, an event could be getting heads in a coin throw or a six in a throw of a dice.

If the occurrence of an event is defined as a success, then

Let the probability of success be denoted by p

Let the probability of non-occurrence of the event or failure be denoted by q

p + q = 1

Let the number of successes be r

And n is the number of trials

Then

**Example:** What are the chances of getting 3 sixes in 10 throws of a dice?

There are 10 trials and 3 events of interest, i.e. successes so:

n = 10

r = 3

The probability of getting a 6 in a dice throw is 1/6, so:

p = 1/6

The probability of not getting a dice throw is:

q = 1 - p = 5/6

P(3 successes) = 10! / ((10 - 3)! 3!) x (5/6)

^{(10 - 3)}x (1/6)^{3}= 10! / (7! x 3!) x (5/6)

^{7}x (1/6)^{3}= 3628800 / (5040 x 6) x (78125 / 279936) x (1/216)

= 0.155

## Winning the Lottery! How to Work out the Odds

We would all like to win the lottery, but the chances of winning are only slightly greater than 0. However "If you're not in, you can't win" and a slim chance is better than none at all!

Take, for example, the California State Lottery. A player must choose 5 numbers between 1 and 69 and 1 Powerball number between 1 and 26. So that is effectively a 5 number selection from 69 numbers *and* a 1 number selection from 1 to 26. To calculate the odds, we need to work out the number of combinations, not permutations, since it doesn't matter what way the numbers are arranged to win.

The number of combinations of r objects is ^{n}C_{r} = n! / ((n - r)! r!)

n = 69

and

r = 5

and

^{n}C_{r}=^{69}C_{5 }= 69! / ( (69 - 5)! 5!) = 69! / (64! 5!) = 11,238,513

So there are 11,238,513 possible ways of picking 5 numbers from a choice of 69 numbers.

Only 1 Powerball number is picked from 26 choices, so there are only 26 ways of doing this.

For every possible combination of 5 numbers from the 69, there are 26 possible Powerball numbers, so to get the total number of combinations, we multiply the two combinations.

So the total possible number of combinations = 11,238,513 x 26 = 292,201,338 or roughly 293 million and the probability of winning is 1 in 293 million.

## Engineering Mathematics by K.A. Stroud - An Excellent Textbook!

Engineering Mathematics by K.A. Stroud is an excellent math textbook for both engineering students and anyone with a general interest in mathematics. The material has been written for part 1 of BSc. Engineering Degrees and Higher National Diploma courses.

A wide range of topics are covered including matrices, vectors, complex numbers, calculus, calculus applications, differential equations and series. The text is written in the style of a personal tutor, guiding the reader through the content, posing questions and encouraging them to provide the answer. Personally, I've found it really easy to follow.

It also covers a more in-depth treatment of probability theory as outlined in this article plus a section on statistics.

This book basically makes learning mathematics fun!

Note: Second hand 1987 editions of this text book are available on Amazon for only about $6

## Did You Find This Article Useful?

Was the info in this article useful and instructive? How can I improve it? Would you like to ask me any questions?

## Questions & Answers

Each Sign has twelve different possibilities, and there are three signs. What are the odds that any two people will share all three signs? Note: the signs can be in different aspects, but at the end of the day each person is sharing three signs. For example, one person could have Pisces as Sun sign, Libra as Rising and Virgo as Moon sign. The other party could have Libra Sun, Pisces Rising, and Virgo moon.

There are twelve possibilities, and each can have three signs = 36 permutations.

But only half of these are a unique combination (e.g., Pisces and Sun is the same as Sun and Pisces)

so that's 18 permutations.

The probability of a person getting one of these arrangements is 1/18

The probability of 2 people sharing all three signs is 1/18 x 1/18 = 1/324

Helpful 3What if someone challenged you to never roll a 3? If you were to roll the dice 18 times, what would be the empirical probability of never getting a three?

The probability of not getting a 3 is 5/6 since there are five ways you can not get a 3 and there are six possible outcomes (probability = no. of ways event can occur / no of possible outcomes). In two trials, the probability of not getting a 3 in the first trial AND not getting a 3 in the second trial (emphasis on the "and") would be 5/6 x 5/6. In 18 trials, you keep multiplying 5/6 by 5/6 so the probability is (5/6)^18 or approximately 0.038.

Helpful 1I have a 12 digit keysafe and would like to know what is the best length to set to open 4,5,6 or 7?

If you mean setting 4,5,6 or 7 digits for the code, 7 digits would of course have the greatest number of permutations.

Helpful 1I am playing a game with 5 possible outcomes. It is assumed that the outcomes are random. For sake of his argument let us call the outcomes 1, 2, 3, 4 and 5. I have played the game 67 times. My outcomes have been: 1 18 times, 2 9 times, 3 zero times, 4 12 times and 5 28 times. I am very frustrated in not getting a 3. What are the odds of not getting a 3 in 67 tries?

Since you carried out 67 trials and the number of 3s was 0, then the empirical probability of getting a 3 is 0/67 = 0, so the probability of not getting a 3 is 1 - 0 = 1.

In a greater number of trials there may be an outcome of a 3 so the odds of not getting a 3 would be less than 1.

Helpful 1If you have nine outcomes and you need three specific numbers to win without repeating a number how many combinations would there be?

It depends on the number of objects n in a set.

In general, if you have n objects in a set and make selections r at a time, the total possible number of combinations or selections is:

nCr = n! / ((n - r)! r!)

In your example, r is 3

Number of trials is 9

The probability of any particular event is 1/nCr and the expectation of the number of wins would be 1/(nCr) x 9.

**© 2016 Eugene Brennan**

## Comments

Thank you Eugene for this tutorial. Very Interesting! Do you recommend any book which goes into more detail, ideally exploring games of chance, sports books etc?

From the following section: What Is the Expectation of an Event?

Why is the answer calculated as 1/6 x 60?

Isn't it the same probability per trial, i.e.:

1st trial = 1/6 chance of getting any number

2nd trial = 1/6 chance of getting any number

and so on...

Therefore, why is it not calculated as (1/6)^60? What am I missing out/confusing, please?

Thanks.

Thank you so much for this article. It was most helpful. It answered questions that bothered me since the days in college!

Wonderful insight into odds.

Thanks for sharing and reiterating the basic mathematics we learn in our early years of schooling! Actually, this topic is very useful in real life even if you engange in a field which does not deal much on numbers such as mine. I agree with Jodah, well-researched hub!

It's nice to know these equations and the odds of throwing certain numbers of dice, drawing a certain card etc. Very well researched hub , Eugene. However under the heading "Probability of an Event" it says; "There are two types of probability, empirical and empirical."(should the second one be "classical"?)

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