Statistics : Chapter 5.5 & 5.6 probability - counting techniques

Multiplication Rule of Counting

If a task consists of a sequence of choices in which there are p selections for the first choice, q selections for the second choice, r selections for the third choice, and so on, then the task of making these selections can be done in p . q . r .......... different ways.

The Factorial DEFINITION

If n >= 0 is an integer, the factorial symbol, n !, is defined as follows: n! = n(n - 1) ... 3 x 2 x 1 remember : 0! = 1 && 1! = 1 For example, 2! = 2 x 1 = 2, 3! = 3 x 2 x 1 = 6, 4! = 4 x 3 x 2 x 1 = 24, and so on.

Solve Counting Problems Using Combinations Definition - Combination :

A combination is a collection, without regard to order, in which r objects are chosen from n distinct objects with r <= n and without repetition. The symbol nCr represents the number of combinations of n distinct objects taken r at a time. In a permutation, order is important. For example, the arrangements ABC and BAC are considered different arrangements of the letters A, B, and C. If order is unimportant, we do not distinguish ABC from BAC. In poker, the order in which the cards are received does not matter. The combination of the cards is what matters.

Example 4: The Travelling Salesperson Problem You have just been hired as a book representative for Pearson Education. On your first day, you must travel to seven schools to introduce yourself. How many different routes are possible?

Approach Call the seven schools A, B, C, D, E, F, and G. School A can be visited first, second, third, fourth, fifth, sixth, or seventh. So, there are seven choices for school A. There are six choices for school B, five choices for school C, and so on. Use the Multiplication Rule and the factorial to find the solution. Solution : 7 x 6 x 5 x 4 x 3 x 2 x 1 = 7! = 5040 different routes are possible.

Combination : Description

Number of Combinations of n Distinct Objects Taken r at a Time . The number of different arrangements of r objects chosen from n objects, in which 1. the n objects are distinct, 2. repetition of objects is not allowed, and 3. order is not important is given by the formula nCr = n!/r!(n - r)! Using Formula (2) to solve the problem presented in Example 7, we obtain 4C2 = 4!/2! x (4 - 2)! =4!/2! x 2! =4 x 3 x 2!/2 x 1 x 2! = 12/2 = 6

Permutations with Nondistinct Items

Example 9 suggests a general result. Had the letters in the sequence each been different, 8P8 = 8! possible sequences would have been formed. This is the numerator of the answer. The presence of two As, two Cs, and three Gs reduces the number of different sequences, as the entries in the denominator illustrate. We are led to the following result: The number of permutations of n objects of which n1 are of one kind, n2 are of a second kind, . . . , and nk are of a kth kind is given by n! /n1! . n2!.... nk! where n = n1 + n2 + .....+ nk .

Formula : Number of Permutations of n Distinct Objects Taken r at a Time : nPr = n!/(n - r)!

The number of arrangements of r objects chosen from n objects, in which 1. the n objects are distinct, 2. repetition of objects is not allowed, and 3. order is important, is given by the formula nPr = n!/(n - r)! So we could represent the solution in Example 3 as nPr = 14P3 = 14 x 13 x 12 = 2184 and the solution to Example 4 as 7P7 = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040

Determine the Appropriate Counting Technique to Use

To determine the appropriate counting technique to use, we need the ability to distinguish between a sequence of choices and an arrangement of items. We also need to determine whether order matters in the arrangements. We first must decide whether we have a sequence of choices or an arrangement of items. For a sequence of choices, we use the Multiplication Rule of Counting if the number of choices at each stage is independent of previous choices. This may involve the rules for arrangements, since each choice in the sequence could involve arranging items. If the number of choices at each stage is not independent of previous choices, we use a tree diagram. When determining the number of arrangements of items, we want to know whether the order of selection matters. If order matters, we also want to know whether we are arranging all the items available or a subset of the items.

Ways to solve the counting problems

solving counting problems : 1) using multiplication rule : 2) using permutations 3) using combinations 4) involving permutations with nondistinct items

Example 7 : Listing Combinations Problem : Roger, Ken, Tom, and Jay are going to play golf. They will randomly select teams of two players each. List all possible team combinations. That is, list all the combinations of the four people Roger, Ken, Tom, and Jay taken two at a time.

Approach : We list the possible teams. We note that order is unimportant, so {Roger,Ken} is the same as {Ken, Roger}. Solution : The list of all such teams (combinations) is Roger, Ken Roger, Tom Roger, Jay Ken, Tom Ken, Jay Tom, Jay So, 4C2 = 6 There are six ways of forming teams of two from a group of four players.

Example 9 :DNA Sequence Problem A DNA sequence consists of a series of letters representing a DNA strand that spells out the genetic code. There are four possible letters (A, C, G, and T). How many distinguishable sequences can be formed using two As, two Cs, three Gs, and one T?

Approach : Each sequence formed will have eight letters. To construct each sequence, we need to fill in eight positions with the eight letters: 1 2 3 4 5 6 7 8 The process of forming a sequence consists of four tasks: Task 1: Choose the positions for the two As. Task 2: Choose the positions for the two Cs. Task 3: Choose the positions for the three Gs. Task 4: Choose the position for the one T. Task 1 can be done in 8C2 ways because we are choosing the 2 positions for A, but order does not matter (because we cannot distinguish the two As).This leaves 6 positions to be filled, so task 2 can be done in 6C2 ways.This leaves 4 positions to be filled, so task 3 can be done in 4C3 ways. The last position can be filled in 1C1 way. Solution : By the Multiplication Rule, the number of possible sequences that can be formed is 8C2 x 6C2 x 4C3 x 1C1 = (8! / 2! x 6!) (6!/2! x 4!) (4!/3! x 1! ) (1! /1! x 0!) = 8! /2! x 2! x 3! x 1! x 0! = 1680

Example 12 : Acceptance Sampling Problem : A shipment of 120 fasteners that contains 4 defective fasteners was sent to a manufacturing plant. The plant's quality-control manager randomly selects and inspects 5 fasteners. What is the probability that exactly 1 of the inspected fasteners is defective?

Approach : Find the probability that exactly 1 fastener is defective by calculating the number of ways of selecting exactly 1 defective fastener in 5 fasteners and dividing this result by the number of ways of selecting 5 fasteners from 120 fasteners. To choose exactly 1 defective in the 5 requires choosing 1 defective from the 4 defectives and 4 nondefectives from the 116 nondefectives. The order in which the fasteners are selected does not matter, so we use combinations. Solution : The number of ways of choosing 1 defective fastener from 4 defective fasteners is 4C1. The number of ways of choosing 4 nondefective fasteners from 116 nondefectives is 116C4. Using the Multiplication Rule, we find that the number of ways of choosing 1 defective and 4 nondefective fasteners is (4C1) x (116C4) = 4 x 7,160,245 = 28,640,980 The number of ways of selecting 5 fasteners from 120 fasteners is 120C5 = 190,578,024. The probability of selecting exactly 1 defective fastener is P(1 defective fastener) = (4C1)(116C4) /120C5 = 28,640,980 /190,578,024 = 0.1503 There is a 15.03% probability of randomly selecting exactly 1 defective fastener.

Multiplication rule : Example 1: Counting the Number of Possible Meals Problem : The fixed-price dinner at a Restaurant provides the following choices: Appetizer: soup or salad Entrée: baked chicken, broiled beef patty, baby beef liver, or roast beef au jus Dessert: ice cream or cheesecake How many different meals can be ordered?

Approach : Ordering such a meal requires three separate decisions: Choose an Appetizer : 2 choices Choose an Entrée : 4 choices Choose a Dessert: 2 choices soup --> chicken soup --> patty soup --> Liver soup --> beef salad --> chicken salad --> patty salad --> Liver salad --> beef with all the 3 possible combinations : soup --> chicken --> ice cream soup --> chicken --> cheesecake soup --> patty --> ice cream soup --> patty --> cheesecake soup --> Liver --> ice cream soup --> Liver --> cheesecake soup --> beef --> ice cream soup --> beef --> cheesecake salad --> chicken --> ice cream salad --> chicken --> cheesecake salad --> patty --> ice cream salad --> patty --> cheesecake salad --> Liver --> ice cream salad --> Liver --> cheesecake salad --> beef --> ice cream salad --> beef --> cheesecake Solution : Look at the tree diagram. For each choice of appetizer, we have 4 choices of entrée, and for each of these 2 x 4 = 8 choices, there are 2 choices for dessert. A total of 2 x 4 x 2 = 16 different meals can be ordered.

Example 8 : Simple Random Samples Problem : How many different simple random samples of size 4 can be obtained from a population whose size is 20?

Approach : The 20 individuals in the population are distinct. so individuals selection order is unimportant. Thus, the number of simple random samples of size 4 from a population of size 20 is a combination of 20 objects taken 4 at a time. Solution Use Formula (2) with n = 20 and r = 4: 20C4 =20!/4!(20 - 4)! =20!/4! x 16! = = 20 x 19 x 18 x 17 x 16! /4 x 3 x 2 x 1 x 16! =116,280/24 = 4845

Compute Probabilities Involving Permutations and Combinations Example 11 : Winning the Lottery Problem : In the Illinois Lottery, an urn contains balls numbered 1 to 52. From this urn, six balls are randomly chosen without replacement. For a $1 bet, a player chooses two sets of six numbers. To win, all six numbers must match those chosen from the urn. The order in which the balls are picked does not matter. What is the probability of winning the lottery?

Approach : The probability of winning is given by the number of ways a ticket could win divided by the size of the sample space. Each ticket has two sets of six numbers and therefore two chances of winning. The size of the sample space S is the number of ways 6 objects can be selected from 52 objects without replacement and without regard to order, so N(S) = 52C6. Solution The size of the sample space is N(S) = 52C6 =52! /6! x (52 - 6)! = 52 x 51 x 50 x 49 x 48 x 47 x 46! /6! x 46! = 20,358,520. Each ticket has two chances of winning. If E is the event "winning ticket," then N(E) = 2 and P(E) = N(E)/N(S) = 2 /20,358,520 = 0.000000098. There is about a 1 in 10,000,000 chance of winning the Illinois Lottery!

Example 3: Counting without Repetition Problem : Three members from a 14-member committee are to be randomly selected to serve as chair, vice-chair, and secretary. The first person selected is the chair, the second is the vice-chair, and the third is the secretary. How many different committee structures are possible?

Approach : The task consists of making three selections. The first selection requires choosing from 14 members. Because a member cannot serve in more than one capacity, the second selection requires choosing from the 13 remaining members. The third selection requires choosing from the 12 remaining members. We use the Multiplication Rule to determine the number of possible committees structures. Solution : By the Multiplication Rule,14 x 13 x 12 = 2184 different committee structures are possible. In this example, repetition is not allowed, unlike Example 2.

Example 2: Counting Airport Codes (Repetition Allowed) Problem : How many different airport codes are possible if we assign three letter codes to represent airport locations.

Approach : We are choosing 3 letters from 26 letters and arranging them in order. Notice that repetition of letters is allowed. Use the Multiplication Rule of Counting, we have 26 ways to choose the first letter, 26 ways to choose the second letter, and 26 ways to choose the third letter. Solution : By the Multiplication Rule,26 x 26 x 26 = 17,576 different airport codes are possible.

Example 10 : Arranging Flags Problem : How many different vertical arrangements are there of 10 flags if 5 are white, 3 are blue, and 2 are red?

Approach : We seek the number of permutations of 10 objects, of which 5 are of one kind (white), 3 are of a second kind (blue), and 2 are of a third kind (red). Solution Using Formula (3), we find that there are 10! /5! x 3! x 2! = 10 x 9 x 8 x 7 x 6 x 5! = 5! x 3! x 2! = 2520 different vertical arrangements

Example 6 :Betting on the horse racing Problem : In how many ways can horses in a 10-horse race finish first, second, and third?

Approach: The 10 horses are distinct. Once a horse crosses the finish line, that horse will not cross the finish line again, and, in a race, finishing order is important. We have a permutation of 10 objects taken 3 at a time. Solution: The top three horses can finish a 10-horse race in 10P3 = 10!/(10 - 3)! =10!/7! =10 x 9 x 8 x 7!/7! = 10 x 9 x 8 = 720 ways

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

In Example 7, if selecting {Roger, Ken} was the same as selecting {Ken, Roger}, so there were 2! = 2 rearrangements of the two objects. This can be determined from the formula for nPr by calculating rPr = r!. So, if we divide nPr by r!, we will have the desired formula for nCr : nCr = nPr/r! =n!/r!(n - r)!

Example 13 : Probability: Which Rule Do I Use? Problem : In the game show Deal or No Deal?, a contestant is presented with 26 suitcases that contain amounts ranging from $0.01 to $1,000,000. The contestant must pick an initial case that is set aside as the game progresses. The amounts are randomly distributed among the suitcases prior to the game as shown in the below table . What is the probability that the contestant picks a case worth at least $100,000? Prize----------------------------- Number of Suitcases $0.01-$100 -------------- -----------8 $200-$1000 ------------------------6 $5,000-$50,000 --------------------5 $100,000-$1,000,000 -------------7

Solution : There is a single event, so we must decide among the empirical, classical, or subjective approaches to determine the probability. The probability experiment is selecting a suitcase. Each prize amount is randomly assigned to one of the 26 suitcases, so the outcomes are equally likely. Table 11 shows that 7 cases contain at least $100,000. Letting E = "worth at least $100,000," we compute P(E) using the classical approach. P(E) = N(E)/N(S) =7/26 = 0.269 The chance the contestant selects a suitcase worth at least $100,000 is 26.9%. In 100 different games, we would expect about 27 games to result in a contestant choosing a suitcase worth at least $100,000.

Example 14 : Probability: Which Rule Do I Use? Problem : According to a Harris poll, 14% of adult Americans have one or more tattoos, 50% have pierced ears, and 65% of those with one or more tattoos also have pierced ears. What is the probability that a randomly selected adult American has one or more tattoos and pierced ears?

Solution : We are finding the probability of an event involving 'AND'. Letting T = "one or more tattoos" and E = "ears pierced," we must find P(T and E). We need to determine if the two events, T and E, are independent. The problem statement tells us that P(E) = 0.50 and P(ET) = 0.65. Because P(E) P(E/T),the two events are not independent. We find P(T and E) using the General Multiplication Rule. P(T and E) = P(T) x P(ET) = (0.14) x (0.65) = 0.091 So the likelihood of selecting an adult American at random who has one or more tattoos and pierced ears is 9.1%.

Example 15 : Counting: Which Technique Do I Use? Problem : A city council consists of 5 men and 4 women. How many different subcommittees can be formed that consist of 3 men and 2 women?

Solution : We need to find the number of subcommittees having 3 men and 2 women. So we consider a sequence of events: select the men, then select the women. Since the number of choices at each stage is independent of previous choices (the men chosen will not impact which women are chosen), we use the Multiplication Rule of Counting to obtain N(subcommittees) = N(ways to pick 3 men) x N(ways to pick 2 women) To select the men, we must consider the number of arrangements of 5 men taken 3 at a time. Since the order of selection does not matter, we use the combination formula. N(ways to pick 3 men) = 5C3 =5!/3! x 2! = 10 To select the women, we must consider the number of arrangements of 4 women taken 2 at a time. Since the order of selection does not matter, we use the combination formula again. N(ways to pick 2 women) = 4C2 =4!/2! x 2! = 6 Combining our results, we obtain N(subcommittees) = 10 x 6 = 60. There are 60 possible subcommittees that contain 3 men and 2 women.

Example 16 : Counting: Which Technique Do I Use? Problem : On February 20, 2011, the Daytona International Speedway hosted the 53rd running of the Daytona 500, touted by many to be the most anticipated event in racing history. With 43 drivers in the race, in how many different ways could the top four finishers (1st, 2nd, 3rd, and 4th place) occur?

Solution : We need to find the number of ways to select the top four finishers. We can view this as a sequence of choices, where the first choice is the first-place driver, the second choice is the second-place driver, and so on. There are 43 ways to pick the first driver,42 ways to pick the second, 41 ways to pick the third, and 40 ways to pick the fourth. The number of choices at each stage is independent of previous choices, so we can use the Multiplication Rule of Counting. The number of ways the top four finishers can occur is N(top four) = 43 x 42 x 41 x 40 = 2,961,840 We could also approach this problem as an arrangement of units. Since each race position is distinguishable, order matters. We are arranging the 43 drivers taken 4 at a time. Using our permutation formula, we get N(top four) = 43P4 =43!/(43 - 4)! =43!/39! = 43 x 42 x 41 x 40 = 2,961,840 Again there are 2,961,840 different ways that the top four finishers could occur.

Determine the Appropriate Probability Rule to Use

The first step is to determine whether we are finding the probability of a single event. If we are dealing with a single event, we must decide whether to use the classical method (equally likely outcomes), the empirical method (relative frequencies), or subjective assignment. For experiments involving more than one event, we first decide which type of statement we have. For events involving 'AND', we must know if the events are independent. For events involving 'OR', we need to know if the events are disjoint (mutually exclusive).