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Chapter 5: Problem 49

Evaluate these probabilities: a. \(\frac{C_{1}^{2} C_{1}^{1}}{C_{2}^{3}}\) b. \(\frac{C_{0}^{4} C_{2}^{2}}{C_{2}^{6}}\) c. \(\frac{C_{2}^{2} C_{1}^{2}}{C_{3}^{4}}\)

Short Answer

Expert verified
Question: What are the values of the following expressions: a. \(\frac{C_{1}^{2} C_{1}^{1}}{C_{2}^{3}}\) b. \(\frac{C_{0}^{4} C_{2}^{2}}{C_{2}^{6}}\) c. \(\frac{C_{2}^{2} C_{1}^{2}}{C_{3}^{4}}\) Answer: a. The value of the expression \(\frac{C_{1}^{2} C_{1}^{1}}{C_{2}^{3}}\) is 0. b. The value of the expression \(\frac{C_{0}^{4} C_{2}^{2}}{C_{2}^{6}}\) is 15. c. The value of the expression \(\frac{C_{2}^{2} C_{1}^{2}}{C_{3}^{4}}\) is 0.

Step by step solution

01

a. Evaluate \(\frac{C_{1}^{2} C_{1}^{1}}{C_{2}^{3}}\)

First, we need to determine the factorials for each combination: - For \(C_1^2\): \(1! = 1\) - For \(C_1^1\): \(1! = 1\) - For \(C_2^3\): \(2! = 2\) Now, we will apply the combination formula for each term: - \(C_1^2 = \frac{1!}{2!(1-2)!} = \frac{1}{2(1)} = 0\) - \(C_1^1 = \frac{1!}{1!(1-1)!} = \frac{1}{1(1)} = 1\) - \(C_2^3 = \frac{2!}{3!(2-3)!} = \frac{2}{6(1)} = \frac{1}{3}\) Finally, we will substitute the values of the combinations into the given expression: \(\frac{C_{1}^{2} C_{1}^{1}}{C_{2}^{3}} = \frac{0 \times 1}{\frac{1}{3}} = 0\)
02

b. Evaluate \(\frac{C_{0}^{4} C_{2}^{2}}{C_{2}^{6}}\)

First, we need to determine the factorials for each combination: - For \(C_0^4\): \(0! = 1\) - For \(C_2^2\): \(2! = 2\) - For \(C_2^6\): \(2! = 2\) Now, we will apply the combination formula for each term: - \(C_0^4 = \frac{1!}{4!(1-4)!} = \frac{1}{24(1)} = \frac{1}{24}\) - \(C_2^2 = \frac{2!}{2!(2-2)!} = \frac{2}{2(1)} = 1\) - \(C_2^6 = \frac{2!}{6!(2-6)!} = \frac{2}{720(1)} = \frac{1}{360}\) Finally, we will substitute the values of the combinations into the given expression: \(\frac{C_{0}^{4} C_{2}^{2}}{C_{2}^{6}} = \frac{\frac{1}{24} \times 1}{\frac{1}{360}} = \frac{1}{24} \times 360 = 15\)
03

c. Evaluate \(\frac{C_{2}^{2} C_{1}^{2}}{C_{3}^{4}}\)

First, we need to determine the factorials for each combination: - For \(C_2^2\): \(2! = 2\) - For \(C_1^2\): \(1! = 1\) - For \(C_3^4\): \(3! = 6\) Now, we will apply the combination formula for each term: - \(C_2^2 = \frac{2!}{2!(2-2)!} = \frac{2}{2(1)} = 1\) - \(C_1^2 = \frac{1!}{2!(1-2)!} = \frac{1}{2(1)} = 0\) - \(C_3^4 = \frac{6!}{4!(3-4)!} = \frac{6}{12(1)} = \frac{1}{2}\) Finally, we will substitute the values of the combinations into the given expression: \(\frac{C_{2}^{2} C_{1}^{2}}{C_{3}^{4}} = \frac{1 \times 0}{\frac{1}{2}} = 0\)

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Key Concepts

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

Factorials
Factorials are an essential part of combinatorics because they are used to determine the number of ways items can be arranged or ordered. A factorial, denoted as \( n! \), is the product of all positive integers up to a specified number \( n \). For example, \( 4! = 4 \times 3 \times 2 \times 1 = 24 \).

Factorials are especially important in calculating permutations and combinations. Here's why:
  • The factorial function helps in determining the likelihood of events by defining the number of possible arrangements or sequences.
  • Factorials are used to eliminate the repetition of permutations when calculating combinations where the sequence does not matter.
  • In probability problems, understanding factorials can simplify the calculation of complex expressions.
Be mindful that for zero factorial, the value is always defined as \( 0! = 1 \). This is a unique property that simplifies many equations in combinatorial mathematics.
Probability Evaluation
Evaluating probabilities often involves understanding the relationships between different combinations and permutations. In the context of probability, combinatorics helps us to count and assess the likelihood of different outcomes.

When evaluating probabilities:
  • Start by defining the total number of possible outcomes using combinations or permutations.
  • Identify the number of successful outcomes that match the criteria of the event.
  • Express the probability as a ratio or a fraction of successful outcomes over the total number of possible outcomes.
The use of factorials and combinations as seen in the provided expressions involves breaking down complex ratios into simpler components. This provides a straightforward path to evaluate which options are more likely to occur. In our original exercise example, we see probability evaluation through dividing combinations to find ratios of likelihood in different scenarios.
Combination Formula
The combination formula is a tool we use to calculate how many ways we can choose items without regard to order from a larger set. It is denoted as \( C(n, r) \) or \( \binom{n}{r} \), where \( n \) is the total number of items, and \( r \) is the number of items to choose.

The formula is: \[ C(n, r) = \frac{n!}{r!(n-r)!} \] This formula calculates how many different groups of `r` items can be formed from a set of `n` items.

Key points:
  • Unlike permutations, combinations do not consider the order of items. Only the selection matters.
  • The combination formula utilizes factorials to divide the total arrangements by the duplications that arise from internal ordering of selected items.
  • This formula is fundamental in areas like binomial probability, lottery calculations, and any scenario involving selection without order.
In our exercise, we apply the combination formula to encapsulate both the concept of choosing items and the likelihood that sequence and order play no role in the outcome, showing its practical importance.

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Most popular questions from this chapter

A company has five applicants for two positions: two women and three men. Suppose that the five applicants are equally qualified and that no preference is given for choosing either gender. Let \(x\) equal the number of women chosen to fill the two positions. a. Write the formula for \(p(x)\), the probability distribution of \(x\). b. What are the mean and variance of this distribution? c. Construct a probability histogram for \(x\).

One model for plant competition assumes that there is a zone of resource depletion around each plant seedling. Depending on the size of the zones and the density of the plants, the zones of resource depletion may overlap with those of other seedlings in the vicinity. When the seeds are randomly dispersed over a wide area, the number of neighbors that a seedling may have usually follows a Poisson distribution with a mean equal to the density of seedlings per unit area. Suppose that the density of seedlings is four per square meter \(\left(\mathrm{m}^{2}\right)\). a. What is the probability that a given seedling has no neighbors within \(1 \mathrm{~m}^{2} ?\) b. What is the probability that a seedling has at most three neighbors per \(\mathrm{m}^{2}\) ? c. What is the probability that a seedling has five or more neighbors per \(\mathrm{m}^{2} ?\) d. Use the fact that the mean and variance of a Poisson random variable are equal to find the proportion of neighbors that would fall into the interval \(\mu \pm 2 \sigma .\) Comment on this result.

A public opinion research firm claims that approximately \(70 \%\) of those sent questionnaires respond by returning the questionnaire. Twenty such questionnaires are sent out, and assume that the president's claim is correct. a. What is the probability that exactly 10 of the questionnaires are filled out and returned? b. What is the probability that at least 12 of the questionnaires are filled out and returned? c. What is the probability that at most 10 of the questionnaires are filled out and returned?

Under what conditions can the Poisson random variable be used to approximate the probabilities associated with the binomial random variable? What application does the Poisson distribution have other than to estimate certain binomial probabilities?

College campuses are graying! According to a recent article, one in four college students is aged 30 or older. Assume that the \(25 \%\) figure is accurate, that your college is representative of colleges at large, and that you sample \(n=200\) students, recording \(x\), the number of students age 30 or older. a. What are the mean and standard deviation of \(x ?\) b. If there are 35 students in your sample who are age 30 or older, would you be willing to assume that the \(25 \%\) figure is representative of your campus? Explain.
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