Topic 2 - Counting AND Probability PDF

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Norwaziah Mahmud 1 STA150 – Topic 2 TOPIC 2: COUNTING AND PROBABILITY Objectives: After completing this chapter, you should be able to  Find the number of outcomes in a sequence of events, using fundamental counting rule.  Find the number of ways that r objects can be selected from n objects, using the permutation rule.  Find the number of ways that r objects can be selected from n objects without regard to order, using the combination rule.  Find the probability of an event, using counting rules.  Determine sample spaces and find the probability of an event, using classical probability or empirical probability.  Find the probability of compound events, using the addition rules.  Find the probability of compound events, using the multiplication rules.  Find the conditional probability of an event. 1

INTRODUCTION: COUNTING RULES Many times a person must know the number of all possible outcomes for a sequence of events. To determine this number, three rules can be used: the fundamental counting rule, the permutation rule, and the combination rule. For example, in our daily life we come across many situations wherein selection and arrangements of objects are necessary. For example, a boy wants to buy 3 different books from a Book Shop. In that shop there are 5 different books. He selects 3 books out of 5. Selection of 3 out of 5 is a combination. He arranges these books in his shelf in different ways every day. Each way of arrangement of books is s permutation. These rules are explained here, and they will be used to find probabilities of events. The first rule is called the fundamental counting rule.

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THE FUNDAMENTAL COUNTING RULE  In a sequences of n events in which the first one has k 1 possibilities and the second events has k 2 and the third has k 3 , and so forth, the total number of possibilities of the sequence will be k1 . k 2 .k 3 …k n In this case and means to multiply.

Norwaziah Mahmud 2 STA150 – Topic 2 Example 1: A paint manufacturer wishes to manufacture several different paints. The categories include Color Type Texture Use

Red, blue, white, black, green, brown, yellow Latex, oil Flat, semi gloss, high gloss Outdoor, indoor

How many different kinds of paint can be made if you can select one color, one type, one texture, and one use? The total no. of possible different paints is: 7*2*3*2=84 ways

Example 2: There are four blood types, A, B, AB, and O. blood can also be Rh+ and Rh-. Finally, a blood donor can be classified as either male or female. How many different ways can a donor have his or her blood labeled? (Ans:16)

Norwaziah Mahmud 3 STA150 – Topic 2 When determining the number of different possibilities of a sequence of events, you must know whether repetitions are permissible. Example 3: The manager of a department store chain wishes to make four-digit identification cards for her employees. a) How many different cards can be made if she uses the digits 1, 2, 3, 4, 5 and 6 and repetition are permitted? Pay attention to the conditions -

Four digit numbers Using 1,2,3,4,5 and 6 only Repetition are permitted

Steps: 1. Make four spaces

2. Determine possible number that can be placed into the space

1/2/3/4/5/6

1/2/3/4/5/6

1/2/3/4/5/6

1/2/3/4/5/6

3. All six digits can be placed because the number can be used more than once. 4.

6

*

6

*

=1296 identification card can be generated.

6

*

6

Norwaziah Mahmud 4 STA150 – Topic 2 b) How many different cards can be made if she uses the digits 1, 2, 3, 4, 5 and 6 and repetition are not permitted? Pay attention to the conditions -

Four digit numbers Using 1,2,3,4,5 and 6 only Repetition are not permitted

Steps: 1. Make four spaces

2. Determine possible number that can be placed into the space

1/2/3/4/5/6

2/3/4/5/6

4/5/6

3/4/5/6

3. For first place all the digits can be used, for the second place the digit decrease by one because the number cannot be repeated. Same goes for third and fourth place. 4.

6

5

=360 identification card can be generated.

4

3

Norwaziah Mahmud 5 STA150 – Topic 2 Example 4: How many three-digit odd numbers can be formed from the digits 0, 1, 2, 3, 4, 5 and 6 if a) Repetition is not allowed? Conditions -

three-digit odd numbers 0, 1, 2, 3, 4, 5 and 6 Repetition is not allowed

Step: 1. Make three spaces.

2. Since the question ask to form an odd number, therefore we must consider last place first because odd number will have odd digit at the end of the number. So there 3 digits can be fill which are 1 or 3 or 5

1

1

Or

1

3 Or

Norwaziah Mahmud 6 STA150 – Topic 2

1

5 3. Since the digit cannot be repeated, the first place can be place with 5 digit. Note that digit 0 cannot be fill in first place because the number generated cannot start with zero (for example: 013 is not 3 digit number)

5

5

1

0/3/4/5/6

2/3/4/5/6

=2

1

Or

5

5

1/2/4/5/6

1

0/2/4/5/6

=25

3

Or

5

1/2/3/4/6

5

0/2/3/4/6

1

5

=25

Norwaziah Mahmud 7 STA150 – Topic 2 Answer: 25+25+25=75 odd number can be generated. b) Repetition is allowed? Conditions -

three-digit odd numbers 0, 1, 2, 3, 4, 5 and 6 Repetition is allowed

Step: 1. Make three spaces.

2. Since the question ask to form an odd number, therefore we must consider last place first because odd number will have odd digit at the end of the number. So there 3 digits can be fill which are 1 or 3 or 5

1

1 Or

1

3

Norwaziah Mahmud 8 STA150 – Topic 2 Or

1

5 3. Since the digit can be repeated, the first place can be place with all digits except zero. Note that digit 0 cannot be fill in first place because the number generated cannot start with zero (for example: 013 is not 3 digit number)

6

7

1

0/1/2/3/4/5/6

1/2/3/4/5/6

= 42

1

Or

6

1/2/3/4/5/6

7

0/1/2/3/4/5/6

1

=42

3

Norwaziah Mahmud 9 STA150 – Topic 2 Or

6

1/2/3/4/5/6

7

0/1/2/3/4/5/6

1

=42

5

Answer: 42+42+42= 126 odd number can be generated. These examples illustrate the fundamental counting rule. In summary: if repetitions are permitted, then the numbers stay the same going from left to right. If repetitions are not permitted, then the numbers decrease by 1 for each place left to right. Two other rules that can be used to determine the total number of possibilities of a sequence of events are permutation rule and the combination rule. 3

FACTORIAL NOTATION These rules use factorial notation. The factorial notation uses the exclamation point. 5! = 5.4.3.2.1 9! = 9.8.7.6.5.4.3.2.1 To use the formulas in the permutation and combination rules, a special definition of 0! is needed. 0! = 1

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PERMUTATIONS Permutation is an arrangement of n objects in a specific order. Example 5: Suppose a business owner has a choice of 5 locations in which to establish her business. She decides to rank each location according to certain criteria, such a price of the store and parking facilities. How many different ways can she rank the 5 locations? 5! = 120 or 5*4*3*2*1=120

Norwaziah Mahmud 10 STA150 – Topic 2 Example 6: Suppose the business owner in Example 4.5 wishes to rank only the top 3 of the 5 locations. How many different ways can she rank them? 5

P3 = 50 or 5*4*3 = 60

Example 7: In how many different ways can the five starting players of a basketball team be introduced to the public? (Ans:120)

Example 8: In how many ways can letters from the word “FRIEND” be arranged if no letter can be used more than once? (Ans:720)

Permutation rule 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 and the formula is n

Pr =

n! (n − r )!

for r = 0, 1, 2,n.

For example, 6 P4 =

6! ( 6 − 4 )!

or

6! 6 .5. 4.3 . 2.1 = = 360 2! 2.1

Example 9: Evaluate each of these. a)

b)

c)

d)

7

P5

12

P4

6

P0

5

P5

n

Pr ,

Norwaziah Mahmud 11 STA150 – Topic 2 Example 10: A television news director wishes to use 3 news stories on an evening show. One story will be the lead story, one will be the second story, and the last will be a closing story. If the director has a total of 8 stories to choose from, how many possible ways can the program be set up? (Ans: 336)

Example 11: A school musical director can select 2 musical plays to present next year. One will be presented in the fall, and one will be presented in the spring. If she has 9 to pick from, how many different possibilities are there? (Ans: 72)

Example 12: Four names are drawn from among the 24 members of a club for the offices of president, vice president, treasurer and secretary. In how many way different ways can this be done? (Ans: 225024)

 The number of permutations of n distinct objects arranged in a circle is (n-1)!

Norwaziah Mahmud 12 STA150 – Topic 2 Example 13: In how many ways can you arrange 8 customers at a table? (8-1)!=5040

Example 14: How many circular permutations are there of four persons playing bridge? (Ans: 6)

Partition Rule •

The number of permutation of n objects of which n1 are of one kind, n 2 are of a second kind,….., nk are of a kth kind, and n1 + n 2 +  nk = n is

n! n 1!×n 2!×n k ! Example 15: How many different permutations can be formed from the word PROBABILITY? Step 1: Count how many letters in the PROBABILITY word. 11 letters Step 2: Identify repeated letters B – 2x I – 2x Step 3:

Norwaziah Mahmud 13 STA150 – Topic 2

Example 16: How many different permutations can be formed from the word SYSTEMATIC? (Ans: 907200)

Example 17: In how many ways can the letters of the word 'STATISTICS' be arranged? (Ans: 50400)

Norwaziah Mahmud 14 STA150 – Topic 2 Example 18: In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together? This question has a condition: vowels always come together

Norwaziah Mahmud 15 STA150 – Topic 2

Step 4:

5! 7! × = 20 ×2520 =50400 3! 2! Example 18a: Additional question. In how many different ways can the letters of the word 'CORPORATION' be arranged so that the consonants always come together? (Ans: 432000)

Example 19: In how many ways can the letters of the word 'LEADER' be arranged? (Ans: 360)

Norwaziah Mahmud 16 STA150 – Topic 2 Example 20: In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together? (Ans: 720)

Example 21: Given six digits 1, 2, 3, 4, 5 and 6. a) Determine the number of four-digit numbers that can be formed using digits given if no repetition is allowed. (Ans: 360)

b) How many four-digit numbers greater than 4000 can be formed using the digits given without repetition? Conditions:    

four-digit numbers 1, 2, 3, 4, 5 and 6. greater than 4000 without repetition

Consider the first place first because the condition is require the number to be more than 4000. So the number that can be placed in the first place is 4, 5 or 6 only.

1

4

5

4

1/2/3/5/6

2/3/5/6

3

3/5/6

= 60

Norwaziah Mahmud 17 STA150 – Topic 2 or

1

4

5

5

1/2/3/4/6

2/3/4/6

3

=60

3/4/6

or

1

5

6

1/2/3/4/5

4

2/3/4/5

3

=60

3/4/5

Answer = 60+60+60=180

Example 22: Given six digits 0, 1, 2, 3, 4, and 5. a) Determine the number of four-digit numbers that can be formed using digits given if no repetition is allowed. (Ans: 300)

b) How many four-digit numbers greater than 4000 can be formed using the digits given without repetition? (Ans: 120)

Norwaziah Mahmud 18 STA150 – Topic 2 Example 23: How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated? (Ans: 20)

Example 24: How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed? (Ans: 5040)

Example 25: If all letters of the word MATHEMATICS arrange in a line. a) How many arrangements are possible if the word must begin with M? Condition: -

must begin with M

MM

ATHEATICS

2! 9! × = 90720 2! 2!× 2! b) What is the probability that all consonants must be put together? This question ask you to find the probability. First, you must know the probability formula:

P ( A) =

n( A) n(S )

Arrangement in event A

Arrangement without condition

Norwaziah Mahmud 19 STA150 – Topic 2 Step 1: Find the arrangement in event A. Event A = all consonants must be put together

7! 5! × = 75600 2!× 2! 2!

n(A)=

Step 2: Find the arrangement without condition n(S)=

11! = 4989600 2!× 2!× 2!

Step 3:

P (A ) =

75600 = 0.0152 4989600

Example 26: In how many different ways can the letters of the word 'OPTICAL' be arranged so that the consonants always come together? (Ans: 576)

Norwaziah Mahmud 20 STA150 – Topic 2 5

COMBINATIONS The differnce between a permutation and a combination is that in a combination, the order or arrangement of the objects is not importants; by contrast, order is important in a permutation.

Example 27: Given the letters A, B, C and D, list the permutations and combinations for selecting two letters. Permutations are

Combinations are

Combination Rule  A combination is the number of ways in selecting r objects from n distinct objects where the order is not important. The number used to determine the number of combinations is as follows. n

Cr =

n! r! (n − r )!

Example 28: Evaluate each expression. a)

5

C2

b)

8

C3

c)

9

C7

d)

3

C0

Norwaziah Mahmud 21 STA150 – Topic 2 Example 29: How many ways can a student choose four out of six questions are in an examination? 6

C4 = 15

Example 30: A newspaper editor has received 8 books to review. He decides that he can use 3 reviews in his newspaper. How many different ways can these 3 reviews be selected? (Ans: 56)

Example 31: In a club there are 7 women and 5 men. A committee of 3 women and 2 men is to be chosen. How many different possibilities are there?

C3 × 5C2 = 350

7

Women

Men

Example 4-32: From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done? Condition: at least 3 men are there on the committee We must consider for men first. Here means we have 3 possible selection for men 3, 4 or 5

Norwaziah Mahmud 22 STA150 – Topic 2 Men 7

C3

Women ×

6

C2

525

6

C1

210

6

C0

21

Or 7

C4

×

Or 7

C5

×

Total

756

Example 33: Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed? (Ans: 210)

Example 34: In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there? (Ans: 209)

Example 35: In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women? (Ans: 11760)

Norwaziah Mahmud 23 STA150 – Topic 2 Example 36: A team of 5 is to be formed from 8 female students and 7 male students. How many ways different teams can be made if? a) There is no restriction. (Ans: 3003)

b) There should be more males than females in the team. (Ans: 1260)

Norwaziah Mahmud 24 STA150 – Topic 2 6

PROBABILITY AND COUNTING RULES The counting rules can be combined with the probability rules to solve many types of probability problems. By using the fundamental counting rule, the permutation rule and the combination rule, you can compute the probability of outcomes of many experiments. Example 4-37: A box contains 24 transistors, 4 of which are defective. If 4 are sold at random, find the following probabilities. This question ask you to find the probability. First, you must know the probability formula:

n(S)= 24 C4 = 10626 20 transistors are non-defective 4 transistors are defective

a) Exactly 2 are defective. A = Exactly 2 are defective n(A)= 4C2 × 20 C2 = 1140

P (A ) =

1140 = 0.1073 10626

b) None is defective. B = None is defective n(B)=4C 0 × 20C 4 = 4845

P (B ) =

4845 = 0.4560 10626

Norwaziah Mahmud 25 STA150 – Topic 2 c) All are defective. C = All are defective n(C)= 4C4 × 20C0 = 1

1 = 9.41× 10 −5 10626 d) At least 1 is defective. D = At least 1 is defective P (C ) =

4

C1 × 20 C3 = 4560

4

C2 × 20 C2 = 1140

n(D)= 4 C3 × 20 C1 = 80 4

C4 × 20 C0 = 1

Total=5781 5781 P (D ) = = 0.5440 10626

Norwaziah Mahmud 26 STA150 – Topic 2 Example 38: In a company there are 7 executives: 4 women and 3 men. Three are selected to attend a management seminar. Find these probabilities. a) All 3 selected will be women. (Ans: 0.1143) b) All 3 selected will be men. (Ans: 0.02857) c) 2 men and 1 woman will be selected. (Ans: 0.3429) d) 1 man and 2 women will be selected. (Ans: 0.5143)

Norwaziah Mahmud 27 STA150 – Topic 2 Example 39: A family is being arranged in a line for a group photograph. If the family consists of a mother, a father, a baby, and five children, find number of ways they can be arranged if the line must begin and end with a parent. (Ans: 1440)

Example 40: There are ten people available for appointment to a committee consisting of six people. In how many ways can the committee be formed if Kamal and Julia must be on the committee? (Ans: 70)

Example 41: If all the letters ...