This section covers permutations and combinations. Arranging Objects The number of ways of arranging n unlike objects in a line is n! (pronounced ‘n factorial’). n! = n × (n – 1) × (n – 2) ×…× 3 × 2 × 1 Example How many different ways can the letters P, Q, R, S be arranged? The answer is 4! = 24. This is because there are four spaces to be filled: _, _, _, _ The first space can be filled by any one of the four letters. The second space can be filled by any of the remaining 3 letters. The third space can be filled by any of the 2 remaining letters and the final space must be filled by the one remaining letter. The total number of possible arrangements is therefore 4 × 3 × 2 × 1 = 4!
n! . Example In how many ways can the letters in the word: STATISTICS be arranged? There are 3 S’s, 2 I’s and 3 T’s in this word, therefore, the number of ways of arranging the letters are: 10!=50 400 Rings and Roundabouts
When clockwise and anti-clockwise arrangements are the same, the number of ways is ½ (n – 1)! Example Ten people go to a party. How many different ways can they be seated? Anti-clockwise and clockwise arrangements are the same. Therefore, the total number of ways is ½ (10-1)! = 181 440 Combinations The number of ways of selecting r objects from n unlike objects is:  Example There are 10 balls in a bag numbered from 1 to 10. Three balls are selected at random. How many different ways are there of selecting the three balls? 10C3 =10!=10 × 9 × 8= 120 Permutations A permutation is an ordered arrangement.
nPr = n! . Example In the Match of the Day’s goal of the month competition, you had to pick the top 3 goals out of 10. Since the order is important, it is the permutation formula which we use. 10P3 =10! = 720 There are therefore 720 different ways of picking the top three goals. Probability The above facts can be used to help solve problems in probability. Example In the National Lottery, 6 numbers are chosen from 49. You win if the 6 balls you pick match the six balls selected by the machine. What is the probability of winning the National Lottery? The number of ways of choosing 6 numbers from 49 is 49C6 = 13 983 816 . Therefore the probability of winning the lottery is 1/13983816 = 0.000 000 071 5 (3sf), which is about a 1 in 14 million chance.
Probability is a measure of how likely something is to happen. It is expressed with a number between 0 and 1 ; 0 means "impossible" and 1 means "certain". For example: suppose you are asked to randomly choose a letter of the alphabet between J and N , inclusive. The possibilities are J , K , L , M , N . The probability that you pick a vowel is 0 . The probability that you pick a consonant is 1 . The probability that you pick K is 1 5 = 0.2 . In general, the probability that an event occurs is given by: P ( event ) = number of favorable outcomes number of possible outcomes
Example: Suppose you roll a six-sided die. Find the probability that you roll a number greater than four. There are six possible outcomes: 1 , 2 , 3 , 4 , 5 and 6 . There are two favorable outcomes: 5 and 6 . So the probability is: P ( greater than 4 ) = 2 6 = 1 3 This is a theoretical probability , as opposed to experimental probability , which is the observed number of favorable outcomes out of a certain number of trials. For instance, suppose you rolled the six-sided die 5 times, and got the following results: 2 , 6 , 4 , 5 , 6 . Then the experimental probability of rolling a number greater than four would be 3 6 or one-half. As the number of trials increases, the experimental probability will become close to the theoretical probability. Calculating probabilities of more than one event can be challenging. Sometime we add the probabilities of each event, and sometime we multiply them. We can use key words (e.g., or, and, etc.) to identify the operations to use. That is, if you have the key words, “either”, “or”, “at least” or their synonyms, you need to add the probabilities of each independent events to find the combined probability. When you have the key words such as “and”, “both”, “all” or their synonyms, you need to multiply the probabilities of each independent event to find the combined probability. For example, when a fair die is thrown, what is the probability of getting a 1 or a 4 ? Here, there are 6 possible outcomes and probability of getting the number 1 or 4 is 1 6 each. So, the probability of getting a 1 or 4 is 1 6 + 1 6 or 2 6 = 1 3 . Now consider choosing a white first and then a green ball from a box of 4 black, 7 green and 9 white identical balls if you replace the ball after the first draw. There are 9 white balls out of 20 identical balls in the box. So, the probability of choosing a white ball first is 9 20 . The ball is replaced after the first draw, so the total number of balls remains the same and there are 7 green balls in the box. So, the probability of choosing a green ball is 7 20 . Now, to find the probability of choosing a white first and then a green ball, you need to multiply the probabilities of the two events. Therefore, the probability of the combined event is 9 20 ⋅ 7 20 or 63 400 . You can also check addition rule in probability and multiplication rule in probability .
Key Terms o Counting problem o Replacement o Permutation o Combination Objectives o Understand how to relate counting of outcomes to probability o Calculate the number of outcomes of a random experiment using permutations and combinations o Know how sampling with or without replacement affects a counting problem When solving more complicated probability problems, we may need to consider series of random experiments or experiments that involve several different aspects, such as drawing two cards from a deck or rolling several dice. In such cases, the ability to calculate relative frequencies (and thus probabilities) requires counting the number of possible outcomes of the experiment. Although counting the number of possible outcomes for simple random experiments, such as the flip of a coin (heads or tails), may be quite simple, counting the number of possible outcomes for the ordered selection of three cards from a standard deck may not be. We will show you how to handle these more complicated counting problems, thereby expanding your ability to solve a range of statistics problems. Replacement and Ordering One crucial aspect of many counting problems in probability is whether replacement is used when sampling. For instance, we may be inquiring about whether a drawing of a name from a hat is statistically "fair"; to do so might require performing a series of trials where a name is pulled from the hat. The question arises as to whether that name should be returned to the hat after being drawn: if it is returned, this selection is called sampling with replacement; if not, then it is called sampling without replacement. In the example case of drawing a name from a hat, the drawn name would be returned to the hat for the following trial so that the conditions of each drawing are the same--thus, this example would involve sampling with replacement. Another aspect of counting problems is the applicability of order for the results of the experiment or experiments. In some cases, we may be interested in an ordered set of results: for example, the number of possible combinations for a padlock. On the other hand, we might be interested in a set of results for which order is not relevant, such as how many five-card hands can be dealt from a standard deck. Order is thus another aspect of counting problems that you should keep in mind when solving them. Permutations Let's say we have a set of n objects (such as playing cards, for instance), and we want to make k selections, with replacement, while taking note of the order. We can derive a formula for this case by imagining how we would go about such an experiment. Because we return the object after each selection, every trial has n objects and therefore n potential outcomes. In the first trial, there are n possible outcomes. In the second trial, there are likewise n possible outcomes, resulting in n(n) = n2 different outcomes for the two successive trials. Following this pattern, if we make k selections with replacement from n objects, there are nk possible outcomes, where the order of the outcome is important. Practice Problem: An opaque bag holds five tiles marked with the letters A, B, C, D, E. If three random selections are made from this bag (with the tile being returned to the bag after each selection), how many different words can be formed? (Assume that a "word" need not be a standard English word.) Solution: This problem involves sampling with replacement, and the order of the results is important (although DEA and ADE both use the same letters, for example, they are not the same word). One approach is to attempt to list each possible word: AAA AAB AAC Interested in learning more? Why not take an online Statistics course? and so on. This approach, although conceptually correct, is lengthy and tedious. Let's use the approach discussed above. In each drawing, we have five potential results: A, B, C, D, or E. Since we are drawing a tile three successive times, the number of possible outcomes is 53, or 125. Obviously, writing out all the possible words would be time-consuming indeed! Our counting method saves us a significant amount of time, however. For the case of k selections from n objects, without replacement but still taking note of order, we want to calculate the number of permutations. We can use the following formula, where the number of permutations of n objects taken k at a time is written as nPk. The factorial notation (!) is also defined below.
(x – 1) (x – 2) . 2 1 To understand how we get this formula, first consider the case where we want to find how many ways we can order all n objects. The first choice allows us n options, the second choice allows us n – 1 options, the third choice allows us n – 2 options, and so on, all the way down to 1. The total number of possible orderings is the product of all these numbers, which we can write as n!. If we only make k selections, then we must choose from n objects, then n – 1 objects, and so on, down to n – k + 1. But this just leads us to the formula for permutations given above, which is illustrated in the following practice problem. Practice Problem: A group of five horses is racing at a track. How many different ways can the horses place in the top three? Solution: We can calculate the number of ways the horses can place in the top three by calculating the number of permutations. Let's also consider the problem from a more fundamental perspective. For first place, there are five different possible horses. For each of these possibilities, there are four remaining possibilities for second place-we then multiply five and four. For third place, we have three remaining possibilities for each of the preceding results--calculate the product of five, four, and three, which is 60. We don't care about the last two places. Note how the formula for permutations is related to our fundamental approach to the problem:
Thus, the result is 60. Combinations Some problems require us to calculate a number of possible outcomes without respect to ordering. Such a case is a lottery drawing where all that is required to win is to pick the correct numbers in any order. This problem requires us to make a change to the formulas above so as to disregard cases that have the same set of objects, but with a different ordering. Consider the case where selection is made without replacement. The number of permutations, nPk, uses the formula given above. We can alter this formula to disregard ordering by eliminating each ordering of each set of objects. Since we are choosing k objects from a set of n objects, those k objects can be ordered in k! ways (see our previous discussion). So, if we simply divide nPk by k!, we then have the number of ways we can select k objects from n without replacement and without regard for order. This is called the number of combinations of n taken k at a time, which is sometimes written
Practice Problem: There are five remaining cards from a standard deck. A player must draw two of them. How many different hands can he draw? Solution: This problem requires us to calculate the number of combinations of five cards taken two at a time. Note that order is unimportant and that drawing the cards cannot involve replacement. We can take the exhaustive and tedious approach of writing out all the possibilities as follows, where we label the cards A, B, C, D, and E. AB BC CD DE AC BD CE AD BE AE This approach indicates that there are 10 possible combinations of 5 cards taken 2 at a time. If we use the combinations formula, we get the same result.
If the problem required us to calculate a much larger number (such as if the player had to select 2 cards from a full deck of 52), then writing out all the possibilities would be unduly time consuming. The formula, however, works in every case. Although we will not consider the case of combinations with replacement in depth, the formula can be shown to be Relating Probability and Counting Now that we can count the number of possible outcomes of various types of random experiments, we can also calculate the relative frequencies (and therefore probabilities) of certain events. To do so (assuming each outcome is equally probable, which is not always the case), simply divide the number of outcomes in the event of interest by the total number of possible outcomes. Thus, if we want to calculate the probability of drawing an ace from a standard deck of playing cards, we can divide the number of outcomes in the event where an ace is drawn (4) by the total number of possible outcomes where any card is drawn (52). The probability is then 1/13. Thus, the counting skills discussed above allow us to calculate the probabilities associated with a variety of problems. Practice Problem: A certain lottery has a hat with the numbers 1 through 10 each written on a single scrap of paper. Three numbers are successively pulled from the hat and set aside in no particular order. If a player can pick four numbers, what is the probability that he wins the lottery? Solution: First, consider the selection of the winning numbers in the lottery. We know that the numbers are not replaced after they are chosen and that the order of the selection is not important; thus, the winning lottery numbers are three unique numbers. Now, we want to calculate a player's chance of winning the lottery if he is able to pick four numbers. The winning combination includes the three numbers chosen in the lottery as well as an additional fourth number, which can be any of the remaining numbers. Apart from the three winning numbers, there are seven other numbers that can be chosen for the fourth number. As a result, the player has seven possible winning combinations. To calculate the probability of winning, we must now find out how many total combinations of 4 numbers can be chosen from 10; to do so, we can use the combinations formula
The probability P of the player winning is thus
|
What is the number of possible outcomes of picking one letter at a time from the word probability?
Pos Terkait
Copyright © 2024 ketiadaan Inc.
