© D.J.Dunn www.freestudy.co.uk 1

MATHEMATICS FOR ENGINEERS

STATISTICS

TUTORIAL 3 – PROBABILITY BASICS

CONTENTS

Probability

Permutations

Combinations

Events

Multiple Events

Sample Space

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1. PROBABILITY, COMBINATIONS AND PERMUTATIONS

Consider the simple case of tossing a fair coin and guessing heads or tails. The tossing

is an EVENT and there are only two possible outcomes – heads or tails.

A SAMPLE SPACE is the set of all possible outcomes. Consider the case of rolling

a six sided dice and guessing the number that is on top. This is a single event with

six possible outcomes, all equally likely. The sample space is 1, 2, 3, 4, 5 and 6.

Suppose that we are guessing if the number on top is an even number. This is a

single event with three possible outcomes 2, 4 and 6 and this is the sample space.

The PROBABILITY of an event is decided by the number of possible outcomes but the probability

of each outcome has to be the same. For example the probability ‘p’ of getting any number with a

six sided dice is p = 1/6.

If we conduct the event more than once, either by repeating them or doing several at the same time,

we have MULTIPLE EVENTS. So long as the probability of an event is the same each time, the

probability of doing it more than once is governed by the MULTIPLICATION RULE.

For example, the probability of predicting two throws of the dice correctly is:

P = p x p = (1/6)(1/6) = 1/36 = 1/62

Note that this means you would need 36 different guesses to cover all possible results and this is the

NUMBER OF PERMUTATIONS. The probability of predicting ‘n’ throws correctly is 1/6n.

In general, if there are ‘1/p’ possible outcomes for each event and there are ‘n’ events then the

probability of getting all correct is P =(1/p)n. The inverse of the probability is the number of

permutations ‘m‘ possible. Number of Permutations m = 1/P = pn

WORKED EXAMPLE No. 1

What is the probability of predicting the outcome of tossing a coin correctly

three times in succession?

SOLUTION

The probability of calling correctly when a coin is tossed is p =1/2.

The probability of getting it correct three times is P = (1/2)(1/2) (1/2) = 1/23= 1/8

In other words three events so n = 3 and two possibilities for each so 1/p = 2 so P = (1/2)3 = 1/8

There is no guarantee that you will ever get a single prediction correct in a case like this but if

you made 8 different predictions you would be sure of getting one right. The number of

permutations is 8.

The next example explains what happens when the probability is not the same for each event.

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WORKED EXAMPLE No. 2

5 balls are numbered 1 to 5 and they are drawn randomly one at a time.

What is the probability of predicting the order in which they are removed?

SOLUTION

The probability of getting the first number correct is 1/5

The probability of getting the second number is 1/4 since there are only 4 left.

The probability of getting the third number is 1/3.

The probability of getting the fourth number is 1/2.

The probability of getting the fifth number is 1/1.

The probability of getting all five numbers is 1/5 x 1/4 x 1/3 x 1/2 x 1/1 = 1/120

You would need 120 permutations to be certain of getting one right.

In other words the number of permutations is simply factorial 5 or 5!

WORKED EXAMPLE No. 3

How many guesses would you need to be sure of correctly predicting the results of four football

matches? Each match can be a win, lose or draw.

SOLUTION

There are three possible results or outcomes 1/p = 3

There are four games (events) so n = 4

The numbers of permutations are m = (1/p) 4 = 3

4 = 3

4 = 3

4 = 81

Alternatively:

The probability of getting one predictions correct is p = 1/3

The probability of getting all of them correct is (1/3) (1/3) (1/3) (1/3) = 1/81

Hence to be sure of getting all four correct you would have to make (1/p)n = 81 permutations.

WORKED EXAMPLE No. 4

What is the probability of getting six lottery balls correct in any order in which they are drawn

if they are numbered 1 to 50 and not replaced after they are drawn?

SOLUTION

The probability of getting one right is 1/50

The probability of getting the second right is 1/48 and so on.

The probability of getting 6 right is hence:

Pr = (1/50) x (1/49) ......(1/45) = 1/11441304000

You would need 11441304000 permutations to be sure of winning.

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Consider this example again. The solution was an example of a simple arithmetical series. Suppose

we want the number of permutations for predicting the correct order of r balls out of a total of n. We

could write this as: n

Pr = n(n-1)(n-2)......{n – (r-1)} n

Pr = n(n-1)(n-2)......{n – r +1)

This may be simplified to two factorials as follows. nPr = n!/(n-r)!

nPr = 50!/(50 – 6)! = 50!/44! = 11441304000

Try this on your calculator using the factorial button. This can apply this to any other thing where n

is the number of different things in total and r is the number to be used to form a group.

An even quicker way with your calculator is to use the function nPr if it has one (it’s a shift function

on mine). Enter 50 press the button and enter 6 followed by = and you get the answer.

COMBINATIONS

Consider the following example of one combination of coloured balls.

WORKED EXAMPLE No. 5

How many permutations can be arranged from a group of three coloured balls (Red, Green and

Yellow or 1, 2 and 3)?

SOLUTION

The permutations represent all the possible ways that the things

can be arranged in one combination and this has been done

graphically giving m = 6 .

We could do with a more mathematical way of solving m. We can see from the arrangement that we

have two possible permutations with red (No.1) first, two with Green (No.2) first and two with

Yellow (No.3) first. There were three sub groups each with two permutations giving a total of 6.

This result is also obtained from factorial 3. m = 3! = (3)(2)(1) = 6

If there were 4 balls we would have four sub groups each with six permutations giving a total of 24. This result is also obtained from factorial 4. m = 4! = (4)(3)(2)(1) = 24

For n objects in a combination the number of permutations m = n!

Let the number of sub-groups be x.

The total permutations are M = x m = xn!

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MULTIPLE PERMUTATIONS

We must now consider what happens when we have x combinations each consisting of n objects.

The following example explains these definitions.

WORKED EXAMPLE No. 6

How many permutations and how many sub-groups can be arranged from two groups of three

coloured balls (Red, Green and Yellow)?

SOLUTION

The sub-groups are shown horizontally instead of vertically as before.

Each sub group has m = 6 permutations.

Clearly there are M =12 in total.

Alternatively xn! = 2 (3)(2) = 12

Next consider the case of only drawing three balls from a total of four.

WORKED EXAMPLE No. 7

How many permutations can be made by drawing any three balls from four combinations of

four balls.

If the three drawn are 1, 2 and 3 then we have 6 permutations.

If the three drawn are 1, 2 and 4 we have another 6 permutations.

If the three drawn are 3, 1 and 4 we have another 6 permutations.

If the three drawn are 2, 3 and 4 we have another 6 permutations.

There are four possible combinations each with six permutations giving a total of 24

permutations.

If there are x possible groups the number of permutations are x r! = 4 (3)(2) = 24

If we have a total number of objects n in each combination and each time we draw r, the number of

possible permutations for each combination is nCr from the previous work.

It follows that M = x r!= x nPr and so x =

nPr / r!

This is the number of groups that can be formed by selecting n from a total of r. r!!rn

n!x

nCr may be interpreted as meaning total n Choose r (covered in outcome 1).

© D.J.Dunn www.freestudy.co.uk 6

)2)......(11)(r(r)(r2)....(1)r1)(nrr)(n(n

)2)......(11)(n(n)(n

r!!rn

n!Cr

n

In other words the top line is first r factors of n and the bottom line is r!

In the last example 4)(1)2)(3(

)(2)3)(4(Cx 3

4

Again its easy to use the nCr function on your calculator enter n press the appropriate button and

enter r followed by = to get the answer.

WORKED EXAMPLE No. 8

How many sub-groups and how many permutations can you make with 2 coloured balls selected

from 5?

SOLUTION

Note n = 5 r = 2 so the top line is the first 2 factors of 5

10)(1)2)(3(

)3)(4(5)(C2

5 M = x r! = 10 (2)(1) = 20 permutations

WORKED EXAMPLE No. 9

How many sub-groups and permutations can you make with 4 coloured balls from selected from

10?

SOLUTION

n = 10 r = 4 The top line is the first 4 factors of 10

210(4)(3)(2)

(7)(10)(9)(8)C x 4

10 M = x r! = 210 (4)(3)(2) = 5040

SELF ASSESSMENT EXERCISE No.1

1. How many permutations are required to be sure of predicting the outcome of tossing a six sided

die four times in succession? (1296)

2. How many permutations are required to be sure of predicting the outcome of six football games

if the results can be win, lose or draw? (243)

3. A form of lottery draw involves the use of 6 machines containing ten numbered balls in each. A

single ball is drawn from each machine in turn. To win you must predict each ball as it is

drawn. What is the probability of winning? (1 in 1000 000)

4. A room contains 10 chairs and people are sent in one at a time to sit down in them at random.

How many possible combinations are there in which the people can be sat. (3628800)

5. How many ways can you arrange cards in groups of 3 from a pack of 52? (132 600)