Lectures prepared by:Mr. K.O. Olorede

Statistics for Physical SciencesStat 203

HARMATTAN 2016KWASU

Follows Bluman & Montgomery:

Probabilitybooks

2.0 Counting RulesThe solution to many statistical experiments involves being able to count the number of points in a sample space. Counting points can be hard, tedious, or both. Fortunately, there are ways to make the counting task easier. This lesson focuses on three rules of counting that can save both time and effort: combinations, permutations, and event multiples.

2.1 Experiment: any process that generates well-defined outcomes.

Example:

Experiment Outcomes Toss a coin Head (H), or Tail (T) Roll a dice 1,2,3,4,5,6 Play a football game Win, Lose, Tie Contest in an election Win, Lose Sit an exam Pass, Fail Rain tomorrow Rain, No rain

2.1.2 Sample Space: the set of all experimental outcomes, denoted by S Example:

Experiment Sample Space Toss a coin 𝑆= ሼ𝐻𝑒𝑎𝑑,𝑇𝑎𝑖𝑙ሽ Roll a dice 𝑆= ሼ1,2,3,4,5,6ሽ Play a football game 𝑆= ሼWin,Lose,Tieሽ Contest in an election 𝑆= ሼWin,Loseሽ Sit an exam 𝑆= ሼPass,Failሽ Rain tomorrow 𝑆= ሼRain,No rainሽ

2.2 Multiple Step Experiment:

Example 2.2-1: Counting Rules for If a family has three children, present the rules for counting how many times of having two girls out of the three children. Solution Let B and G denote boys and girls respectively. You can obtain sample space S using a tree diagram as shown below

𝑆=ሼ𝐵𝐵𝐵,𝐵𝐵𝐺,𝐵𝐺𝐵,𝐵𝐺𝐺,𝐺𝐵𝐵,𝐺𝐵𝐺,𝐺𝐺𝐵,𝐺𝐺𝐺ሽ

Example 2.2-2: Counting Rules in Coin and Dice Tossing Experiment A coin is flipped and a die is rolled. Find the sample space. Solution

Example 2.2-3: Possible Choices of Choosing ATM Pin

Example 2.2-3: [Selecting Applicants at Job Interview]

Factorial Arithmetic, Permutation and Combination

2.3 Factorial Arithmetic In 1808, Christian Kramp first used a counting product called the Factorial notation. Factorials are denoted by the exclamation mark (!). The number of different arrangements of n different objects is denoted by 𝑛!. Hence 𝑛!= 𝑛× ሺ𝑛− 1ሻ× ሺ𝑛− 2ሻ×,⋯,× 4× 3× 2× 1 0! = 0 1! = 1 5! = 5× 4× 3× 2× 1 = 120 5! = 5× 4! = 120 4! = 4× 3! = 24 3! = 3× 2! = 6 2! = 2× 1 = 2 etc.

𝑺𝒊𝒎𝒑𝒍𝒊𝒇𝒚 𝒏!ሾ𝟒−(𝒏+𝟑)ሿሺ𝒏−𝟑ሻሺ𝒏−𝟒ሻ!

Solution 𝑛!ሾ4− (𝑛+ 3)ሿሺ𝑛− 3ሻሺ𝑛− 4ሻ! = 𝑛ሺ𝑛− 1ሻሺ𝑛− 2ሻሺ𝑛− 3ሻሺ𝑛− 4ሻ!ሾ4− (𝑛+ 3)ሿ

ሺ𝑛− 3ሻሺ𝑛− 4ሻ!

𝑛!ሾ4−(𝑛+3)ሿሺ𝑛−3ሻሺ𝑛−4ሻ! = 𝑛ሺ𝑛− 1ሻሺ𝑛− 2ሻሾ4− (𝑛+ 3)ሿ

Factorial arithmetic: Problems

𝑆ℎ𝑜𝑤 𝑡ℎ𝑎𝑡 𝑛ቀ 𝑛𝑟− 1ቁ= 𝑟ቀ𝑛𝑟ቁ+ሺ𝑟− 1ሻቀ 𝑛𝑟− 1ቁ

Solution

From R.H.S

𝑟ቀ𝑛𝑟ቁ+ሺ𝑟− 1ሻቀ 𝑛𝑟− 1ቁ= 𝑟ሺ𝑛!ሻሺ𝑛− 𝑟ሻ!𝑟!+ሺ𝑟− 1ሻ 𝑛!

൫𝑛−ሺ𝑟− 1ሻ൯!ሺ𝑟− 1ሻ! 𝑟ቀ𝑛𝑟ቁ+ሺ𝑟− 1ሻቀ 𝑛𝑟− 1ቁ= 𝑟ሺ𝑛!ሻሺ𝑛− 𝑟ሻ!𝑟!+ሺ𝑟− 1ሻ 𝑛!

൫𝑛−ሺ𝑟− 1ሻ൯!ሺ𝑟− 1ሻ! 𝑟ቀ𝑛𝑟ቁ+ሺ𝑟− 1ሻቀ 𝑛𝑟− 1ቁ= 𝑟ሺ𝑛!ሻሺ𝑛− 𝑟ሻ!𝑟!+ሺ𝑟− 1ሻ 𝑛!

ሺ𝑛− 𝑟+ 1ሻ!ሺ𝑟− 1ሻ! 𝑟ቀ𝑛𝑟ቁ+ሺ𝑟− 1ሻቀ 𝑛𝑟− 1ቁ= 𝑟ሺ𝑛!ሻሺ𝑛− 𝑟ሻ!𝑟!+ሺ𝑟− 1ሻ 𝑛!

ሺ𝑛− 𝑟+ 1ሻሺ𝑛− 𝑟ሻ!ሺ𝑟− 1ሻሺ𝑟− 2ሻ! 𝑟ቀ𝑛𝑟ቁ+ሺ𝑟− 1ሻቀ 𝑛𝑟− 1ቁ= 𝑟ሺ𝑛!ሻሺ𝑛− 𝑟ሻ!𝑟!+ 𝑛!

ሺ𝑛− 𝑟+ 1ሻሺ𝑛− 𝑟ሻ!ሺ𝑟− 2ሻ! 𝑟ቀ𝑛𝑟ቁ+ሺ𝑟− 1ሻቀ 𝑛𝑟− 1ቁ= 𝑛!ሺ𝑛− 𝑟ሻ! 𝑟𝑟!+ 1

ሺ𝑛− 𝑟+ 1ሻሺ𝑟− 2ሻ!൨

𝑟ቀ𝑛𝑟ቁ+ሺ𝑟− 1ሻቀ 𝑛𝑟− 1ቁ= 𝑛!ሺ𝑛− 𝑟ሻ! 𝑟𝑟ሺ𝑟− 1ሻ!+ 1

ሺ𝑛− 𝑟+ 1ሻሺ𝑟− 2ሻ!൨ 𝑟ቀ𝑛𝑟ቁ+ሺ𝑟− 1ሻቀ 𝑛𝑟− 1ቁ= 𝑛!ሺ𝑛− 𝑟ሻ! 1

ሺ𝑟− 1ሻ!+ 1ሺ𝑛− 𝑟+ 1ሻሺ𝑟− 2ሻ!൨ 𝑟ቀ𝑛𝑟ቁ+ሺ𝑟− 1ሻቀ 𝑛𝑟− 1ቁ= 𝑛!

ሺ𝑛− 𝑟ሻ! 1ሺ𝑟− 1ሻሺ𝑟− 2ሻ!+ 1

ሺ𝑛− 𝑟+ 1ሻሺ𝑟− 2ሻ!൨ 𝑟ቀ𝑛𝑟ቁ+ሺ𝑟− 1ሻቀ 𝑛𝑟− 1ቁ= 𝑛!ሺ𝑛− 𝑟ሻሺ𝑟− 2ሻ! 1

ሺ𝑟− 1ሻ+ 1ሺ𝑛− 𝑟+ 1ሻ൨ 𝑟ቀ𝑛𝑟ቁ+ሺ𝑟− 1ሻቀ 𝑛𝑟− 1ቁ= 𝑛!

ሺ𝑛− 𝑟ሻሺ𝑟− 2ሻ! 𝑛− 𝑟+ 1+ 𝑟− 1ሺ𝑟− 1ሻሺ𝑛− 𝑟+ 1ሻ൨ 𝑟ቀ𝑛𝑟ቁ+ሺ𝑟− 1ሻቀ 𝑛𝑟− 1ቁ= 𝑛!

ሺ𝑛− 𝑟ሻሺ𝑟− 2ሻ! 𝑛ሺ𝑟− 1ሻሺ𝑛− 𝑟+ 1ሻ൨ 𝑟ቀ𝑛𝑟ቁ+ሺ𝑟− 1ሻቀ 𝑛𝑟− 1ቁ= 𝑛!𝑛

ሺ𝑛− 𝑟+ 1ሻ!ሺ𝑟− 1ሻ! 𝑟ቀ𝑛𝑟ቁ+ሺ𝑟− 1ሻቀ 𝑛𝑟− 1ቁ= 𝑛ቀ 𝑛𝑟− 1ቁ Hence

ቀ𝑛𝑟− 1ቁ= 𝑟ቀ𝑛𝑟ቁ+ሺ𝑟− 1ሻቀ 𝑛𝑟− 1ቁ 𝑎𝑠 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑

Show that 𝑛𝐶𝑟 = 𝑛𝐶𝑛−𝑟 Solution 𝑛𝐶𝑟 = 𝑛𝐶𝑛−𝑟 Recall that 𝑛𝐶𝑟 = 𝑛!

ሺ𝑛−𝑟ሻ!𝑟! and 𝑛𝐶𝑛−𝑟 = 𝑛!൫𝑛−ሺ𝑛−𝑟ሻ൯!ሺ𝑛−𝑟ሻ!

Hence, from R.H.S 𝑛𝐶𝑛−𝑟 = 𝑛!൫𝑛−ሺ𝑛− 𝑟ሻ൯!ሺ𝑛− 𝑟ሻ! 𝑛𝐶𝑛−𝑟 = 𝑛!ሺ𝑛− 𝑛+ 𝑟ሻ!ሺ𝑛− 𝑟ሻ! 𝑛𝐶𝑛−𝑟 = 𝑛!𝑟!ሺ𝑛− 𝑟ሻ! ≡ 𝑛!

ሺ𝑛− 𝑟ሻ!𝑟! 𝐿.𝐻.𝑆.= 𝑅.𝐻.𝑆 𝑎𝑠 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑

Show that 𝒏𝑪𝒓 = 𝒏𝑷𝒓𝒓!

Solution

𝑹𝒆𝒄𝒂𝒍𝒍 𝒕𝒉𝒂𝒕 𝒏𝑪𝒓 = 𝒏!ሺ𝒏− 𝒓ሻ!𝒓! 𝒂𝒏𝒅 𝒏𝑷𝒓 = 𝒏!

ሺ𝒏− 𝒓ሻ! From L.H.S

𝑛𝐶𝑟 = 𝑛!ሺ𝑛− 𝑟ሻ!𝑟!

𝑛𝐶𝑟 = 𝑛!ሺ𝑛− 𝑟ሻ!× 1𝑟!

𝑛𝐶𝑟 = 𝑛𝑃𝑟 × 1𝑟! 𝑛𝐶𝑟 = 𝑛𝑃𝑟𝑟! 𝑎𝑠 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑

Factorize 3× 12!− 10! Solution 3× 12!− 10! = 3× 12× 11× 10!− 10! 3× 12!− 10! = 10!ሺ3× 12× 11− 1ሻ

𝑺𝒊𝒎𝒑𝒍𝒊𝒇𝒚 ሺ𝒏− 𝟐ሻ!𝟑ሺ𝒏!ሻ

Solution ሺ𝑛− 2ሻ!3ሺ𝑛!ሻ = ሺ𝑛− 2ሻ!3𝑛ሺ𝑛− 1ሻሺ𝑛− 2ሻ! ሺ𝑛−2ሻ!3ሺ𝑛!ሻ = 13𝑛ሺ𝑛−1ሻ

5.1 Permutation Permutation simply means arrangement in mathematical terms. It is an arrangement of n objects in a specified order. Therefore, the arrangement of n objects in a specific order using r objects at a time is called a permutation of n objects taking r objects at a time. It is written as 𝑛𝑃𝑟 = 𝑛!

ሺ𝑛−𝑟ሻ!. 5𝑃3 = 5!

ሺ5−3ሻ! = 5!2! = 5×4×3×2!2! = 5× 4× 3 = 60 5𝑃5 = 5!

ሺ5− 5ሻ! = 5!0! = 5× 4× 3× 2× 11 = 120

20𝑃4 = 20!ሺ20 − 4ሻ! = 20!16! = 20× 19× 18× 17× 16!16! = 20× 19× 18× 17= 116280

Example1-5: [Arrangement of graduands at a convocation

ceremony]Suppose you to arrange graduands from the 6 colleges in KWASU for collection of certificates at a convocation ceremony taking only 2 colleges at a time. Determine the number of possible ways. Solution Any college can be selected first in 6 ways, any other college can be selected in the remaining (6-1) ways. The required number of ways = 6∙5 = 30𝑤𝑎𝑦𝑠 Alternatively, 𝑛 = 6 𝑎𝑛𝑑 𝑟= 2 6𝑃2 = 6!

ሺ6−2ሻ! = 6!4! = 6×5×4!4! = 30𝑤𝑎𝑦𝑠

Example 3-5: [Manufacturing Tests]

An inspector must select 3 tests to perform in a certain order on a manufactured part. He has a choice of 7 tests. How many ways can he perform 3 different tests? 210 Solution 7𝑃3 = 7!

ሺ7− 3ሻ!= 7!4! = 7× 6× 5× 4!4! = 210 𝑤𝑎𝑦𝑠

Example 4-5:[Arranging answers in a test paper]

A test paper has six questions but four are to be answered. In how many ways can the answers be arranged? Solution First question can be answered in 6different ways, second can be answered in 5 different ways, third can be answered in 4 different ways and the fourth can be answered in 3 different ways. The required number of ways is therefore 6∙5∙4∙3 = 360𝑤𝑎𝑦𝑠 Alternatively, 6𝑃4 = 6!

ሺ6−4ሻ! = 6!2! = 6×5×4×3×2!2! = 30𝑤𝑎𝑦𝑠

5.1.1 Permutation (arrangement) of identical objects

The number of ways of arranging n objects taking r at a time of which 𝑟1,𝑟2,⋯,𝑟𝑛 are identical is 𝑛!𝑟1!×𝑟2!,⋯𝑟𝑛! Example 5-5 In how many ways can letters of the word MATHEMATICS be arranged? Solution There are 11 letters of which there 2As, 2Ms and 2Ts. Hence the required number of ways is 11!2!×2!×2! = 4989600𝑤𝑎𝑦𝑠

5.1.2 Conditional PermutationThis involves arrangements of n objects taking r of them a

time following some restrictions on the order (or manner) of the arrangement. For instance we may be interested in the number of ways in which the letters of the word SHALLOW can be arranged if i. the two Ls must not come together ii. the two Ls must always come together

SolutionBy removing the two Ls, the remaining letter (SHAOW) can be arranged in 5! Ways. i. If the two Ls must not come together, the first L can occupy

any of 6 places below •S•H•A•O•W• When this is done, there are 5 places for the second L not next to the first. Hence, the required number of arrangements with separate Ls 5! •6•5 supposing the Ls are distinguishable. But here, the two Ls are nit distinguishable. Hence the number of arrangements is 5! •6•52 = 5!× 15 = 1800𝑤𝑎𝑦𝑠

ii. The two Ls are identical here. we take them as one object. There are 6 places for each of the 5! arrangements of the word SHAOW. Hence, the required number of ways = 5!× 6 = 6! =720 𝑤𝑎𝑦𝑠

Example 6-5Find the number of arrangements using all the letters of the word PERCENTAGE if Es must be placed next to each other. Solution By removing the Es, the word PERCENTAGE becomes PRCNTAG and it can be arranged in 7! Ways. The letter Es are taken as one (since they must be placed side by side) and can occupy the 8 places in •P•R•C•N•T•A•G• Hence, the required number of arrangements 7!× 8 = 8! =40320 𝑎𝑟𝑟𝑎𝑛𝑔𝑒𝑚𝑒𝑛𝑡𝑠

5.1.3 Permutation in ring (or cyclic permutation)

This involves arrangement of n objects in a cycle, in a thread (such as beads in thread) or in a round table (such as people in a round table meeting or conference) such that one of them must be taken as a start point or reference the leaving the remaining (n-1)! for arrangement from either directions. It should be noted that if the objects can move about or rotate in their positions, the number arrangements should be ሺ𝑛−1ሻ!2 .

Example 7-5a. In how many ways can six people take places at a

round table? b. How many ways are there if two people must sit next

to each other? c. How many ways are there if two people must not sit

next o each other? Solution

a. Here, one of them will take a place (no matter where). So, there are 5 choices for the second person, 4 choices for the third person, 3 choices for the fourth person, 2 choices for the fifth person and 1 choice for the sixth. The number of ways = 5•4•3•2•1 = 120

c. For two people to sit next to each other at the table, there are 2 possible ways: the second can sit to the right or left of the first. The third person then has a choice of 4 places, the fourth person has a choice of 3 places, the fifth person has a choice of 2 places and the 6 person has a choice of 1 place. Hence the number of ways = 2•4•3•2•1 = 48

d. Two methods are possible here, Method1 Number of ways of sitting 6 people on round table = 120

Figure 1-5: Permutation at a round table

Number of ways of sitting 6 people, if two people must sit next to each other = 48. Hence number of ways of sitting 6 people at a round table if 2 people must not sit next to each other = 120 -48 = 72

C

•

D

•E

•

F

A

Method 2:Refer to figure 1-5 above. Let one of the two people (say A) take a position, leaving behind the other person for the moment. The rest people (C,D,E,F) can take their places in 4! Ways. By now B has 3 choices so as not to sit next to A. Therefore, the required number of ways = 24× 3 = 72 𝑤𝑎𝑦𝑠 Exercise 1-5

1. In how many ways can four men and two women be seated at a round table if the women do not sit next to each other.