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Chapter 2: Problem 30

A friend of mine is giving a dinner party. His current wine supply includes 8 bottles of zinfandel, 10 of merlot, and 12 of cabernet (he only drinks red wine), all from different wineries. a. If he wants to serve 3 bottles of zinfandel and serving order is important, how many ways are there to do this? b. If 6 bottles of wine are to be randomly selected from the 30 for serving, how many ways are there to do this? c. If 6 bottles are randomly selected, how many ways are there to obtain two bottles of each variety? d. If 6 bottles are randomly selected, what is the probability that this results in two bottles of each variety being chosen? e. If 6 bottles are randomly selected, what is the probability that all of them are the same variety?

Short Answer

Expert verified
a. 336 ways; b. 593775 ways; c. 83160 ways; d. Probability 鈮� 0.140; e. Probability 鈮� 0.002.

Step by step solution

01

Calculate ways to serve 3 bottles of zinfandel

The order of serving is important, so we need to use permutations. We have 8 bottles of zinfandel and need to choose 3, respecting the order. The number of ways to pick them is given by the permutation formula: \( P(n, k) = \frac{n!}{(n-k)!} \), where \( n = 8 \) and \( k = 3 \). Therefore, \( P(8, 3) = \frac{8!}{(8-3)!} = \frac{8 \times 7 \times 6}{1} = 336 \).
02

Determine ways to select 6 bottles from 30

We need to choose 6 bottles out of the total 30 without regard for order, so we use combinations. This is calculated as \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \( n = 30 \) and \( k = 6 \). Thus, \( \binom{30}{6} = \frac{30!}{6! \times 24!} = 593775 \).
03

Calculate ways to obtain two bottles of each variety

We select 2 bottles each of zinfandel, merlot, and cabernet. This involves combinations for each type and multiplying the results. For zinfandels: \( \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28 \). For merlots: \( \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \). For cabernets: \( \binom{12}{2} = \frac{12 \times 11}{2 \times 1} = 66 \). Total ways are \( 28 \times 45 \times 66 = 83160 \).
04

Compute probability of two bottles of each variety

Probability is the ratio of the number of favorable outcomes to the total. We've already calculated these as 83160 ways for two bottles of each and 593775 total ways to choose any 6 bottles. Therefore, the probability is \( \frac{83160}{593775} \approx 0.140 \).
05

Determine probability of selecting 6 bottles of the same variety

Calculate probabilities separately for each variety and sum them. For zinfandel: \( \binom{8}{6} = 28 \). For merlot: \( \binom{10}{6} = 210 \). For cabernet: \( \binom{12}{6} = 924 \). Total ways = 28 + 210 + 924 = 1162. Probability is \( \frac{1162}{593775} \approx 0.002 \).

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Key Concepts

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

Combinatorics
Combinatorics is a branch of mathematics that focuses on counting, arrangement, and selection of objects. It's especially helpful when predicting the number of possible outcomes in a situation involving choices. In other words, it's all about figuring out how many ways we can organize, select, or arrange items.

In our dinner party example, combinatorics helps us determine the number of ways wine bottles can be picked or arranged. Whether you care about the order or not, combinatorics gives a framework to approach these problems. This field is divided into subtopics like permutations and combinations, which we'll delve into in the following sections. They are essential tools for evaluating different scenarios involving counting, each with its own specific formulas and rules.
Permutations
Permutations refer to the arrangements of objects in a specific order. The order matters here, which makes permutations different from combinations.

If you're arranging books on a shelf, the sequence counts as different permutations of those books. When we calculate the number of permutations, we use the formula for permutations of \( n \) objects taken \( k \) at a time, which is:
  • \( P(n, k) = \frac{n!}{(n-k)!} \)
In our wine example, serving 3 bottles of zinfandel in a particular order uses permutations. We have 8 bottles, and we need to arrange 3, giving us \( P(8, 3) = 336 \) possible arrangements. This quantifies how many ways we can order the bottles if order is crucial.
Combinations
Combinations involve selecting items from a group, where order does not matter. This is crucial when the order of selection isn't important. Think of it like choosing a team from a larger group without caring about the order of team members, which is different from permutations where order is key.

The formula to find combinations is:
  • \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)
In the dinner party situation, selecting 6 bottles out of 30 without caring about order is a combination problem. You calculate it as \( \binom{30}{6} = 593775 \), which tells us the number of ways to pick the bottles when only the selection matters, not the sequence.
Probability Calculations
Probability in mathematics provides a measure of how likely an event is to occur. It's often expressed as a fraction between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.

To calculate the probability of an event, you divide the number of favorable outcomes by the total number of possible outcomes. In our wine example, we might be interested in picking 2 bottles of each wine variety. This involves calculating the favorable ways (83160) divided by the total possible ways to pick 6 bottles (593775), giving us approximately 0.140 or 14%.

This simple formula helps us assess different scenarios, including the probability of choosing 6 bottles all of the same variety, further emphasizing how combinatorics, permutations, and combinations interplay to solve complex real-world problems relating to chances and likelihoods.

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Most popular questions from this chapter

Seventy percent of the light aircraft that disappear while in flight in a certain country are subsequently discovered. Of the aircraft that are discovered, \(60 \%\) have an emergency locator, whereas \(90 \%\) of the aircraft not discovered do not have such a locator. Suppose a light aircraft has disappeared. a. If it has an emergency locator, what is the probability that it will not be discovered? b. If it does not have an emergency locator, what is the probability that it will be discovered?

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Consider the type of clothes dryer (gas or electric) purchased by each of five different customers at a certain store. a. If the probability that at most one of these purchases an electric dryer is \(.428\), what is the probability that at least two purchase an electric dryer? b. If \(P\) (all five purchase gas) \(=.116\) and \(P\) (all five purchase electric) \(=.005\), what is the probability that at least one of each type is purchased?

A certain system can experience three different types of defects. Let \(A_{i}(i=1,2,3)\) denote the event that the system has a defect of type \(i\). Suppose that $$ \begin{aligned} &P\left(A_{1}\right)=.12 \quad P\left(A_{2}\right)=.07 \quad P\left(A_{3}\right)=.05 \\ &P\left(A_{1} \cup A_{2}\right)=.13 \quad P\left(A_{1} \cup A_{3}\right)=.14 \\\ &P\left(A_{2} \cup A_{3}\right)=.10 \quad P\left(A_{1} \cap A_{2} \cap A_{3}\right)=.01 \end{aligned} $$ a. What is the probability that the system does not have a type 1 defect? b. What is the probability that the system has both type 1 and type 2 defects? c. What is the probability that the system has both type 1 and type 2 defects but not a type 3 defect? d. What is the probability that the system has at most two of these defects?

Human visual inspection of solder joints on printed circuit boards can be very subjective. Part of the problem stems from the numerous types of solder defects (e.g., pad nonwetting, knee visibility, voids) and even the degree to which a joint possesses one or more of these defects. Consequently, even highly trained inspectors can disagree on the disposition of a particular joint. In one batch of 10,000 joints, inspector A found 724 that were judged defective, inspector B found 751 such joints, and 1159 of the joints were judged defective by at least one of the inspectors. Suppose that one of the 10,000 joints is randomly selected. a. What is the probability that the selected joint was judged to be defective by neither of the two inspectors? b. What is the probability that the selected joint was judged to be defective by inspector \(B\) but not by inspector A?
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