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A chili recipe calls for seven ingredients: ground beef, onions, beans, tomatoes, peppers, chili powder, and salt. There are no directions about the order in which the ingredients should be combined. You decide to add the ingredients in a random order. a. How many different arrangements are there? b. What is the probability that onions are first? c. What is the probability that the order is exactly as listed above? d. What is the probability that the order isn't exactly as listed above? e. What is the probability that beans are third?

Short Answer

Expert verified
a: 5040 b: 1/7 c: 1/5040 d: 5039/5040 e: 1/7

Step by step solution

01

Calculate Total Arrangements

The total number of arrangements of the seven ingredients can be found by calculating the factorial of the number of ingredients. Therefore, the total number of arrangements is given by the formula: \[ 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040. \]
02

Probability that Onions are First

If onions are first, we only need to arrange the remaining 6 ingredients. Therefore, the number of such arrangements is \( 6! = 720 \). The probability that onions are first is then the fraction of these arrangements out of the total arrangements:\[ P(\text{onions first}) = \frac{720}{5040} = \frac{1}{7}. \]
03

Probability of Original Order

There is only 1 arrangement that matches the original list exactly. Therefore, the probability of the exact order is:\[ P(\text{exact order}) = \frac{1}{5040}. \]
04

Probability that Order isn't Exact

The probability that the order isn't exactly as listed is simply 1 minus the probability of the exact order:\[ P(\text{not exact order}) = 1 - \frac{1}{5040} = \frac{5039}{5040}. \]
05

Probability that Beans are Third

With beans in the third position, we must arrange the remaining 6 ingredients in the remaining positions. So, the number of arrangements with beans third is \( 6! = 720 \). The probability that beans are third is:\[ P(\text{beans third}) = \frac{720}{5040} = \frac{1}{7}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factorial
When dealing with problems related to the arrangement of items, the factorial function is an indispensable tool. A factorial, denoted by an exclamation mark (!), is used to calculate the number of ways in which a set of items can be arranged. For any positive integer \( n \), the factorial (\( n! \)) is the product of all positive integers less than or equal to \( n \). For example, \( 7! \) is calculated as follows:
  • Start with the number, 7.
  • Then, multiply it by 6, then by 5, continuing down to 1.
  • So, \( 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \).
The factorial function grows very quickly with larger numbers; as such, it’s particularly useful for computing permutations and combinations, especially when calculating the total possible arrangements of a set of items. In the context of our example with 7 ingredients for a chili recipe, the factorial function helps us determine there are 5040 different potential orders in which these ingredients can be added.
Combinatorics
Combinatorics is a branch of mathematics that focuses on counting, arrangement, and combination of elements within a certain set. It's important to understand this concept when calculating probabilities related to arrangements. In our chili recipe example, we are using combinatorial principles to figure out how many different ways the ingredients can be ordered.
  • For determining different orders of adding 7 ingredients, the factorial concept \( 7! \) indicates all possible permutations.
  • Permutations consider the specific order or sequence, so changing the order changes the permutation.
In Scenario b, the problem is to find the probability that onions are first among 7 ingredients:
  • If onions are to be first, then the remaining 6 ingredients need to be arranged afterwards, which makes it \( 6! \) ways, or 720 permutations.
  • The probability is the specific order (onions first) divided by the total possible orders, which is \( \frac{720}{5040} = \frac{1}{7} \).
Combinatorics deftly assists in solving these kinds of problems by applying counting to determine possible configurations.
Arrangement Calculation
Arrangement calculation involves computing the number of ways to order items when the order is important. Returning to our exercise, calculating arrangements helps us identify probabilities associated with specific item placements. Here's how it is applied to certain probabilities from our chili recipe scenario:
  • To find out the probability that a specific ingredient like beans is third, we keep beans fixed in the third position.
  • Then, arrange the remaining 6 ingredients (which is \( 6! = 720 \) ).
  • The probability is then \( \frac{720}{5040} = \frac{1}{7} \), because it’s one specific configuration among the 5040 possible configurations.
Arrangement calculations highlight how specific positions can impact outcomes. Other examples include determining the probability of an exact sequential order, like the original listing.
  • Only one arrangement matches the exact order, hence the probability \( \frac{1}{5040} \).
  • Conversely, the probability of not having the exact original order is \( 1 - \frac{1}{5040} = \frac{5039}{5040} \).
Understanding the method behind these calculations allows for accurate counting and probability determinations in arrangements.

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Most popular questions from this chapter

APPLICATION The star hitter on the baseball team at City Community College had a batting average of 375 before the start of a three-game series. (Note: Batting average is calculated by dividing hits by times at bat; sacrifice bunts and walks do not count as times at bat.) During the three games, he came to the plate to bat eleven times. In these eleven plate appearances, he walked twice and had one sacrifice bunt. He either got a hit or struck out in his other plate appearances. If his batting average was the same at the end of the three-game series as at the beginning how many hits did he get?

A class has 7 male students and 14 female students. One student is selected at random from the class, and then a second student is selected from the class. Draw a tree diagram that shows the events "male" and "female" for the two consecutive selections. Write probabilities on the branches. (a)

The tree diagram of outcomes for rolling two dice would have 36 equally likely paths. It is easier to look at a situation like this using a two-way table instead of a tree diagram. a. Complete a table like this showing the sums of two six-sided dice. b. Calculate the expected sum of rolling two dice. (h)

State whether each pair of events is dependent or independent. a. Roll a die, then roll the same die again. b. Remove one card from the deck, then draw a second card. (a) c. Flip a coin, then flip a second coin.

In a random walk, you move according to rules with each move being determined by a random process. The simplest type of random walk is a one-dimensional walk where each move is either one step forward or one step backward on a number line. a. Start at 0 on the number line and flip a coin to determine your move. Heads means you take one step forward to the next integer, and tails means you take one step backward to the previous integer. What sequence of six tosses will land you on the number-line locations \(+1,+2,+1,+2,+3,+2\) ? (a) b. Explore a one-dimensional walk of 100 moves using a calculator routine that randomly generates \(+1\) or \(-1\). In list \(\mathrm{L} 1\), generate random numbers with 1 representing a step forward and \(-1\) representing a step backward. Describe what you need to do with list L1 to show your number-line location after every step. [r \([\) See Calculator Note 10A and 10B. 4] c. Describe the results of your simulation. Is this what you expected?

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