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Chapter 4: Problem 13

A food company plans to conduct an experiment to compare its brand of tea with that of two competitors. A single person is hired to taste and rank each of three brands of tea, which are unmarked except for identifying symbols \(A, B\), and \(C\). a. Define the experiment. b. List the simple events in \(S\). c. If the taster has no ability to distinguish a difference in taste among teas, what is the probability that the taster will rank tea type \(A\) as the most desirable? As the least desirable?

Short Answer

Expert verified
Answer: The probability that the taster will rank tea type A as the most desirable is \(\frac{1}{3}\), and the probability that the taster will rank tea type A as the least desirable is also \(\frac{1}{3}\).

Step by step solution

01

Define the experiment

The experiment consists of a single person tasting and ranking three brands of tea labeled as A, B, and C. The taster will rank the teas in order of preference, with no ties in ranking allowed.
02

List the simple events in S

There are 3! (3 factorial) possible ways to rank the three teas, as each rank can be occupied by any one of the three types. So there are 3 options for the first rank, 2 for the second rank, and 1 for the third rank. The simple events in S are the possible rankings of the teas: {(A, B, C), (A, C, B), (B, A, C), (B, C, A), (C, A, B), (C, B, A)}.
03

Calculate the probability that A is ranked most desirable

If the taster cannot distinguish any difference in taste among teas, each of the 6 possibilities is equally likely. To find the probability that A is ranked as the most desirable, we need to count how many of the simple events have A in the top rank. There are 2 events with A as the most desirable: {(A, B, C), (A, C, B)}. The probability of A being ranked most desirable is \(\frac{2}{6} = \frac{1}{3}\).
04

Calculate the probability that A is ranked least desirable

Similarly, to find the probability that A is ranked as the least desirable, we need to count how many of the simple events have A in the third rank. There are 2 events with A as the least desirable: {(B, C, A), (C, B, A)}. The probability of A being ranked least desirable is \(\frac{2}{6} = \frac{1}{3}\). So, the probability that the taster will rank tea type A as the most desirable is \(\frac{1}{3}\), and the probability that the taster will rank tea type A as the least desirable is also \(\frac{1}{3}\).

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Key Concepts

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

Simple Events
In probability theory, an event is an outcome or set of outcomes from a random experiment. Simple events are the most basic possible outcomes we can identify in an experiment, and they cannot be broken down further. These are individual results from a sample space, which is the set of all possible outcomes.
  • In the tea ranking experiment, each potential order of preference is a simple event.
  • The taster can rank the three teas in various sequences, leading to different simple events.
  • For instance, ranking tees as (A, B, C) or (B, C, A), each represents a different simple event.
Recognizing simple events helps determine overall probabilities and clarify possible results from an experiment.
Factorial Calculation
Factorials are an essential part of counting arrangements and permutations in mathematics. The factorial of a number, denoted by an exclamation mark (!), is the product of all positive integers up to that number.
  • For the ranking of the three types of tea, we need to calculate the total number of possible arrangements, which involves finding the factorial of 3.
  • Calculating 3! means multiplying 3 脳 2 脳 1, resulting in 6 possible arrangements.
  • This calculation is critical because it tells us how many simple events exist in the experiment.
Factorial calculations help in determining the size of the sample space, which is indispensable for calculating probabilities.
Equally Likely Outcomes
When outcomes are equally likely, each possible result has the same probability of occurring in an experiment. This assumption is key in many probability exercises, simplifying calculations.
  • In the tea ranking experiment, if the taster cannot distinguish between the teas, each ranking is equally probable.
  • This results in each of the 6 possible rankings having an equal chance of \( \frac{1}{6} \).
  • Using this concept allows us to easily calculate the probability of any particular event occurring, assuming all are equally likely.
This principle helps simplify understanding potential outcomes by assigning them equal chances in the absence of further distinguishing information.
Ranking Probabilities
When ranking and assigning probabilities, it's important to know how many different rankings correspond to a specific outcome.
  • For example, when looking for the probability of tea A being ranked first, we observe how many rankings begin with A.
  • In our scenario, outcomes (A, B, C) and (A, C, B) satisfy this condition, hence a probability of \( \frac{2}{6} = \frac{1}{3} \).
  • Likewise, for A to be least desirable, outcomes (B, C, A) and (C, B, A) apply, also resulting in \( \frac{1}{3} \).
Understanding the process of identifying and calculating these rankings guides us in deriving conclusions from probability experiments effectively.

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Most popular questions from this chapter

A certain manufactured item is visually inspected by two different inspectors. When a defective item comes through the line, the probability that it gets by the first inspector is .1. Of those that get past the first inspector, the second inspector will "miss" 5 out of \(10 .\) What fraction of the defective items get by both inspectors?

A quality-control plan calls for accepting a large lot of crankshaft bearings if a sample of seven is drawn and none are defective. What is the probability of accepting the lot if none in the lot are defective? If \(1 / 10\) are defective? If \(1 / 2\) are defective?

Suppose that \(P(A)=.3\) and \(P(B)=.5 .\) If events \(A\) and \(B\) are mutually exclusive, find these probabilities: a. \(P(A \cap B)\) b. \(P(A \cup B)\)

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Two stars of the LA Lakers are very different when it comes to making free throws. ESPN.com reports that Kobe Bryant makes \(85 \%\) of his free throw shots while Lamar Odum makes \(62 \%\) of his free throws. \(^{5}\) Assume that the free throws are independent and that each player shoots two free throws during a team practice. a. What is the probability that Kobe makes both of his free throws? b. What is the probability that Lamar makes exactly one of his two free throws? c. What is the probability that Kobe makes both of his free throws and Lamar makes neither of his?
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