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Chapter 6: Problem 54

Three friends \((\mathrm{A}, \mathrm{B}\), and \(\mathrm{C})\) will participate in a round-robin tournament in which each one plays both of the others. Suppose that \(P(\) A beats \(B)=.7, P(\) A beats \(C)=.8\), \(P(\mathrm{~B}\) beats \(\mathrm{C})=.6\), and that the outcomes of the three matches are independent of one another. a. What is the probability that \(\mathrm{A}\) wins both her matches and that B beats C? b. What is the probability that A wins both her matches? c. What is the probability that A loses both her matches? d. What is the probability that each person wins one match? (Hint: There are two different ways for this to happen.)

Short Answer

Expert verified
a) The probability that A wins both matches and B beats C is \(0.7 * 0.8 * 0.6 = 0.336\). b) The probability that A wins both her matches is \(0.7 * 0.8 = 0.56\). c) The probability that A loses both her matches is \(0.3 * 0.2 = 0.06\). d) The probability that each person wins one match is \((0.7 * 0.6 * 0.2) + (0.3 * 0.4 * 0.8) = 0.268\).

Step by step solution

01

Part (a) Solution

The task is to find the probability that A wins both matches and B beats C. Since these events are independent, the combined probability is simply the product of individual probabilities. So take the probability of A winning against B, which is 0.7, multiply it by the probability of A winning against C, which is 0.8, and finally multiply it by the probability of B winning against C, which is 0.6.
02

Part (b) Solution

The task is to find the probability that A wins both her matches. Since the matches are independent, simply multiply the probability of A winning against B, which is 0.7, by the probability of A winning against C, which is 0.8.
03

Part (c) Solution

The task is to find the probability that A loses both her matches. The event of A losing a match is the complement of A winning. So the probability of A losing against B is \(1 - P(B)\), which is 0.3, and against C it's \(1 - P(C)\), which is 0.2. As these are also independent, simply multiply these two probabilities.
04

Part (d) Solution

The task is to find the probability that each person wins one match. There are two cases for this: either A beats B, B beats C, and C beats A, or C beats B, B beats A, and A beats C. Compute the probabilities for each case by multiplying the probabilities of those individual matches, and then add them for the final probability, as these two cases are mutually exclusive and therefore the probabilities can be added.

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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 events whose outcomes do not affect each other. If two events are independent, the probability of both occurring is the product of the probabilities of each occurring individually. For example, if we consider the matches in a round-robin tournament, the outcome of one match does not influence another, assuming independence. This means the probability of A winning against B is independent of B's match with C, or A's match with C.

For independent events:
  • The probability that Event A occurs is denoted by \( P(A) \).
  • The probability that Event B occurs is \( P(B) \).
  • To find the probability that both Event A and Event B occur, multiply them: \( P(A \text{ and } B) = P(A) \cdot P(B) \).
Understanding independence allows us to calculate compound probabilities simply by products, simplifying many problems in statistics and probability.
Round-Robin Tournament
A round-robin tournament is a competition format where each participant plays against every other participant. This can often be seen in sports leagues and games. For example, if three friends are playing against each other, each person will play two matches鈥攐ne against each of the other participants.

In a round-robin tournament:
  • Every participant competes against every other participant exactly once.
  • The total number of games in a round-robin tournament with \( n \) players is \( \frac{n(n-1)}{2} \).
This format helps in ensuring fairness as each player competes with all others under similar conditions. In probability exercises, it's common to calculate the outcome likelihoods using independent events as each game is considered independent from others.
Complement Rule
The complement rule in probability refers to the concept that the probabilities of all possible outcomes of an event add up to 1. Therefore, the probability of an event not occurring is simply 1 minus the probability of the event occurring.

Mathematically, it's expressed as:
  • If \( P(A) \) is the probability that event A occurs, then \( P(A') = 1 - P(A) \) is the probability that event A does not occur.
This rule is particularly useful when it's simpler to calculate the probability of an event not happening. For instance, if we know the probability of A winning against B is 0.7, then the probability of A losing to B is \( 1 - 0.7 = 0.3 \).

Using this rule alongside independent events makes it easier to solve complex probability questions, such as finding out how often a certain undesirable (or complementary) outcome will happen.

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Most popular questions from this chapter

An individual is presented with three different glasses of cola, labeled C, D, and P. He is asked to taste all three and then list them in order of preference. Suppose that the same cola has actually been put into all three glasses. a. What are the simple events in this chance experiment, and what probability would you assign to each one? b. What is the probability that \(\mathrm{C}\) is ranked first? c. What is the probability that \(\mathrm{C}\) is ranked first \(a n d \mathrm{D}\) is ranked last?

A company that manufactures video cameras produces a basic model and a deluxe model. Over the past year, \(40 \%\) of the cameras sold have been the basic model. Of those buying the basic model, \(30 \%\) purchase an extended warranty, whereas \(50 \%\) of all purchasers of the deluxe model buy an extended warranty. If you learn that a randomly selected purchaser bought an extended warranty, what is the probability that he or she has a basic model?

The National Public Radio show Car Talk has a feature called "The Puzzler." Listeners are asked to send in answers to some puzzling questions-usually about cars but sometimes about probability (which, of course, must account for the incredible popularity of the program!). Suppose that for a car question, 800 answers are submitted, of which 50 are correct. Suppose also that the hosts randomly select two answers from those submitted with replacement. a. Calculate the probability that both selected answers are correct. (For purposes of this problem, keep at least five digits to the right of the decimal.) b. Suppose now that the hosts select the answers at random but without replacement. Use conditional probability to evaluate the probability that both answers selected are correct. How does this probability compare to the one computed in Part (a)?

Only \(0.1 \%\) of the individuals in a certain population have a particular disease (an incidence rate of .001). Of those who have the disease, \(95 \%\) test positive when a certain diagnostic test is applied. Of those who do not have the disease, \(90 \%\) test negative when the test is applied. Suppose that an individual from this population is randomly selected and given the test. a. Construct a tree diagram having two first-generation branches, for has disease and doesn't have disease, and two second-generation branches leading out from each of these, for positive test and negative test. Then enter appropriate probabilities on the four branches. b. Use the general multiplication rule to calculate \(P(\) has disease and positive test). c. Calculate \(P\) (positive test). d. Calculate \(P\) (has disease \(\mid\) positive test). Does the result surprise you? Give an intuitive explanation for the size of this probability.

A construction firm bids on two different contracts. Let \(E_{1}\) be the event that the bid on the first contract is successful, and define \(E_{2}\) analogously for the second contract. Suppose that \(P\left(E_{1}\right)=.4\) and \(P\left(E_{2}\right)=.2\) and that \(E_{1}\) and \(E_{2}\) are independent events. a. Calculate the probability that both bids are successful (the probability of the event \(E_{1}\) and \(E_{2}\) ). b. Calculate the probability that neither bid is successful (the probability of the event \(\left(\right.\) not \(\left.E_{1}\right)\) and \(\left(\right.\) not \(\left.E_{2}\right)\) ). c. What is the probability that the firm is successful in at least one of the two bids?
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