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Chapter 14: Problem 16

If the subject guesses two shapes correctly and two incorrectly, in how many ways can you arrange the sequence of correct and incorrect guesses? (a) \((32)=3\left(\begin{array}{l}3 \\ 2\end{array}\right)=3\) (b) \((42)=12\left(\begin{array}{l}4 \\ 2\end{array}\right)=12\) (c) \((42)=6\left(\begin{array}{l}4 \\ 2\end{array}\right)=6\)

Short Answer

Expert verified
The correct answer is option (c) with 6 ways to arrange the guesses.

Step by step solution

01

Understand the problem

The problem is about arranging two correct (C) and two incorrect (I) guesses. We need to figure out in how many ways we can arrange these four guesses into a sequence.
02

Apply the Combination Formula

To determine the number of sequences, we can use the combination formula, which selects positions for the correct guesses. The formula for combinations is \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \), where \(n\) is the total number of items, and \(r\) is the number of items to choose from.
03

Substitute the Values

In this problem, we have 4 positions to fill (2 correct and 2 incorrect), so \(n = 4\) and \(r = 2\). Substitute these values into the combination formula: \( \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \).
04

Verify the Solution

Let's verify: With the combination \( \binom{4}{2} = 6 \), there are 6 unique sequences for two correct and two incorrect guesses, which confirms that answer (c) is correct.

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Key Concepts

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

Permutation and Combination
In order to tackle problems related to arranging or selecting items, two primary concepts in combinatorics are used: permutation and combination.
  • Permutation: This concept is about arranging items in a specific order. Permutations consider the sequence as important, so different orders are counted separately.
  • Combination: This is used when the order does not matter, only the selection or grouping of items is considered. This is perfect for scenarios like selecting team members from a group where the order does not matter.
In the given exercise, we specifically use the combination formula. We are interested in finding how many ways we can choose positions for 2 correct guesses out of 4 total guesses (which naturally determines the positions for the 2 incorrect guesses). Since order does not matter (a 'correct' guess is a 'correct' guess, irrespective of its position), combinations are the right tool here.
Mathematical Formulas
Mathematical formulas are essential for solving combinatorics problems efficiently. In this exercise, the combination formula is the key.The combination formula is given by:\[\binom{n}{r} = \frac{n!}{r!(n-r)!}\]
  • \(n\): Total number of items or positions to consider.
  • \(r\): Number of items or positions we are choosing.
  • \(!\) (factorial): The factorial of a number \(n\) (denoted as \(n!\)) is the product of all positive integers less than or equal to \(n\).
In our problem, \(n = 4\) and \(r = 2\), representing the four positions we are sequence-arranging and the two correct guesses that need placement. Plugging these values into our formula simplifies to:\[\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6\]This tells us that there are 6 different ways to sequence the guesses.
Problem Solving Steps
Solving a problem systematically can enormously help in understanding and arriving at the correct solution.Let's break down the steps applied in this combinatorics exercise:
  • Step 1: Understanding the problem: Here, the task is to arrange guesses where two are correct (C) and two are incorrect (I). Understanding this determines what combinatorial approach to use.
  • Step 2: Applying the combination formula: Having established that order doesn鈥檛 matter, we use the combination formula to determine the arrangements. We know our \(n\) and \(r\), and we calculate \(\binom{4}{2}\).
  • Step 3: Substituting values: Plug into the formula: substituting \(n = 4\), \(r = 2\) gives \(\binom{4}{2} = 6\).
  • Step 4: Verifying the solution: Always verify your result to ensure it makes sense in the context. We find that there are indeed 6 potential ways to arrange C & I in this scenario as checked by re-evaluation.
Following these steps, ensures clarity and correctness in solving similar combinatorial problems.

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Most popular questions from this chapter

Binomial setting? In each of the following situations, is it reasonable to use a b矛nomial distribution for the random variable \(X\) ? Give reasons for your answer in each case. (a) An auto manufacturer chooses one car from each hour's production for a detailed quality inspection. One variable recorded is the count \(X\) of finish defects (dimples, ripples, etc.) in the car's paint. (b) The pool of potential jurors for a murder case contains 100 persons chosen at random from the adult residents of a large city. Each person in the pool is asked whether he or she opposes the death penalty; \(X\) is the number who say "Yes." (c) Joe buys a ticket in his state's Pick 3 lottery game every week; \(X\) is the number of times in a year that he wins a prize.

Binomial setting? A binomial distribution will be approximately correct as a model for one of these two sports settings and not for the other. Explain why by briefly discussing both settings. (a) A National Football League kicker has made \(80 \%\) of his field goal attempts in the past. This season he attempts 20 field goals. The attempts differ widely in distance, angle, wind, and 50 on. (b) A National Basketball Association player has made \(80 \%\) of his free- throw attempts in the past. This season he takes 150 free throws. Basketball free throws are always attempted from 15 feet away from the basket with no interference from other players.

Disabled Adults in Canada. Statistics Canada reports that \(13.7 \%\) of adult Canadians (aged 15 and older) report being limited in their daily activities due to a disability. \({ }^{4}\) If you take an SRS of 4000 Canadians aged 15 and over, what is the approximate distribution of the number in your sample who report being limited in their daily activities due to a disability? Explain why the approximation is valid in this situation.

Aerosolized vaccine for measles. An aerosolized vaccine for measles was developed in Mexico and has been used on more than 4 million childiren since 1980. Aerosolized vaccines have the advantages of being able to be administered by people without clinical training and do not cause injectionassociated infections. The percentage of children developing an immune response to measles after receiving the subcutaneous injection of the vaccine is \(95 \%\), and for those receiving the aerosolized vaccine, it is \(85 \%\). 9 There are 20 children to be vaccinated for measles using the aerosolized vaccine in a small rural village in India. We are going to count the number who developed an immune response to measles after vaccination. (a) Explain why this is a binomial setting. (b) What is the probability that at least one child does not develop an immune response to measles after receiving the aerosolized vaccine? What would be the probability that at least one child does not develop an immune response to measles if all children were vaccinated using the subcutaneous injection of the vaccine?

Estimating \(\pi\) from random numbers. Kenyon College student Eric Newman used basic geometry to evaluate software random number generators as part of a summer research project. He generated 2000 independent random points \((X, Y)\) in the unit square. [That is, \(X\) and \(Y\) are independent random numbers between 0 and 1, each having the density function illustrated in Figure \(12.4\) (page 290). The probability that \((X, Y)\) falls in any region within the unit square is the area of the region. \(]^{14}\) (a) Sketch the unit square, the region of possible values for the point \((X, Y)\). (b) The set of points \((X, Y)\) where \(X^{2}+Y^{2}<1\) describes a circle of radius 1 . Add this circle to your sketch in part (a), and label the intersection of the two regions \(A\). (c) Let \(T\) be the total number of the 2000 points that fall into the region \(A\). \(T\) follows a binomial distribution. Identify \(n\) and \(p\). (Hint: Recall that the area of a circle is \(\pi r^{2}\)-) (d) What are the mean and standard deviation of \(T\) ? (e) Explain how Eric used a random number generator and the facts given here to estimate \(\pi\).
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