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Chapter 2: Problem 98

Each contestant on a quiz show is asked to specify one of six possible categories from which questions will be asked. Suppose \(P(\) contestant requests category \(i)=\frac{1}{6}\) and successive contestants choose their categories independently of one another. If there are three contestants on each show and all three contestants on a particular show select different categories, what is the probability that exactly one has selected category 1 ?

Short Answer

Expert verified
The probability is \( \frac{5}{18} \).

Step by step solution

01

Determine Single Category 1 Selection Odds

Let's determine the probability that exactly one contestant chooses category 1. For contestant 1 to choose category 1, the probability is \( \frac{1}{6} \). For contestant 1 to not choose category 1 but rather one of the remaining 5 categories, the probability is \( \frac{5}{6} \). This needs to hold true for contestants 2 and 3 as well.
02

Consider Permutation of Choices

Since any one of the three contestants can be the one who picks category 1, the number of ways to choose which contestant picks category 1 is 3. Thus, we need to multiply the probability calculated for one specific arrangement by 3.
03

Calculate Probability for Exactly One Choosing Category 1

The probability of one specific contestant picking category 1 and the other two contestants picking different categories is \( \frac{1}{6} \times \left( \frac{5}{6} \right) \times \left( \frac{4}{6} \right) \). This takes into account the requirement that each contestant selects different categories. Simplifying, we get \( \frac{1}{6} \times \frac{5}{6} \times \frac{4}{6} = \frac{20}{216} \).
04

Adjust for Multiple Permutations

To account for any of the three contestants being the one to select category 1, multiply by the number of permutations: \( 3 \). The adjusted probability is \( 3 \times \frac{20}{216} = \frac{60}{216} \).
05

Simplify the Probability

Simplify \( \frac{60}{216} \) by finding the greatest common divisor of 60 and 216. The simplified probability is \( \frac{5}{18} \).

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Key Concepts

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

Independent Events
In probability theory, independent events are those whose occurrence does not affect each other. This means the outcome of one event does not change the probability of the other. In the quiz show example, the category choice of one contestant doesn't impact the choice of another. Each has a \( \frac{1}{6} \) chance of picking any category, regardless of what others pick. This makes calculations simpler because we can multiply individual probabilities.
  • The probability remains the same irrespective of prior outcomes.
  • This is crucial in calculations as it allows modeling successive outcomes as independent events.
  • In scenarios involving multiple participants, this characteristic helps in managing complexity and maintaining uniformity in probability assignments.
Permutation
Permutation deals with determining the number of possible arrangements or orders when selecting items. This is important in our problem because, although three contestants can choose from several categories, only one truly selects category 1. Because any one of the three contestants could be the one who picks category 1, we calculate permutations.
  • The number of ways to select which contestant chooses category 1 is 3, as any one of them can make the choice while others choose differently.
  • This necessitates multiplying the probability of a specific sequence by 3, validating different orders.
  • Permutations allow us to adjust probabilities for all possible sequences, providing more accurate results.
Simplifying Fractions
Simplifying fractions involves reducing them to their lowest terms. To do this, divide both the numerator and denominator by their greatest common divisor (GCD). In the context of our problem, we achieved this when reaching the probability of exactly one contestant selecting category 1.
  • Initially calculated probability was \( \frac{60}{216} \), a more complex fraction.
  • The greatest common divisor of 60 and 216 is 12.
  • Dividing both by 12 simplifies the fraction to \( \frac{5}{18} \), making it more interpretable and cleaned up for final representation.
  • Reducing fractions helps in comparing and understanding final results more clearly.
Probability Calculation
Calculating probability often involves combining different concepts such as multiplication rules and permutations. In this problem, calculating the chance of exactly one contestant picking category 1 involves multiplying the probability that the first contestant chooses category 1, while the others choose differently.
  • First, compute the probability of one specific arrangement: \( \frac{1}{6} \times \frac{5}{6} \times \frac{4}{6} \).
  • This sequence maintains that the remaining two contestants choose different categories.
  • Multiply by 3 to account for each contestant having an equal opportunity to be the one choosing category 1.
  • The final calculation leads to \( \frac{5}{18} \), accounting for all desired conditions.
Understanding these calculations can demystify complex probability problems, turning them into manageable steps through recognized probability rules.

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Most popular questions from this chapter

Consider randomly selecting a single individual and having that person test drive 3 different vehicles. Define events \(A_{1}\), \(\mathrm{A}_{2}\), and \(\mathrm{A}_{3}\) by \(A_{1}=\) likes vehicle \(\\# 1 \quad A_{2}=\) likes vehicle \(\\# 2\) \(A_{3}=\) likes vehicle \(\\# 3\) Suppose that \(P\left(A_{1}\right)=.55, P\left(A_{2}\right)=.65, P\left(A_{3}\right)=.70\), \(P\left(A_{1} \cup A_{2}\right)=.80, P\left(A_{2} \cap A_{3}\right)=.40\), and \(P\left(A_{1} \cup A_{2} \cup A_{3}\right)=.88\). a. What is the probability that the individual likes both vehicle #1 and vehicle #2? b. Determine and interpret \(P\left(A_{2} \mid A_{3}\right)\). c. Are \(\mathrm{A}_{2}\) and \(\mathrm{A}_{3}\) independent events? Answer in two different ways. d. If you learn that the individual did not like vehicle #1, what now is the probability that he/she liked at least one of the other two vehicles?

Show that for any three events \(A, B\), and \(C\) with \(P(C)>0\), \(P(A \cup B \mid C)=P(A \mid C)+P(B \mid C)-P(A \cap B \mid C) .\)

Consider independently rolling two fair dice, one red and the other green. Let \(A\) be the event that the red die shows 3 dots, \(B\) be the event that the green die shows 4 dots, and \(C\) be the event that the total number of dots showing on the two dice is 7. Are these events pairwise independent (i.e., are \(A\) and \(B\) independent events, are \(A\) and \(C\) independent, and are \(B\) and \(C\) independent)? Are the three events mutually independent?

A small manufacturing company will start operating a night shift. There are 20 machinists employed by the company. a. If a night crew consists of 3 machinists, how many different crews are possible? b. If the machinists are ranked \(1,2, \ldots, 20\) in order of competence, how many of these crews would not have the best machinist? c. How many of the crews would have at least 1 of the 10 best machinists? d. If one of these crews is selected at random to work on a particular night, what is the probability that the best machinist will not work that night?

Computer keyboard failures can be attributed to electrical defects or mechanical defects. A repair facility currently has 25 failed keyboards, 6 of which have electrical defects and 19 of which have mechanical defects. a. How many ways are there to randomly select 5 of these keyboards for a thorough inspection (without regard to order)? b. In how many ways can a sample of 5 keyboards be selected so that exactly two have an electrical defect? c. If a sample of 5 keyboards is randomly selected, what is the probability that at least 4 of these will have a mechanical defect?
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