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Chapter 6: Problem 12

Three friends \((A, B,\) and \(C)\) will participate in a round-robin tournament in which each one plays both of the others. Suppose that \(P(\mathrm{~A}\) beats \(\mathrm{B})=.7, P(\mathrm{~A}\) beats C) \(=.8\), and \(P(B\) beats \(C)=.6\) and that the outcomes of the three matches are independent of one another. a. What is the probability that \(A\) wins both her matches and that \(\mathrm{B}\) beats \(\mathrm{C}\) ? 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. Calculate the probability of each separately, and then add.)

Short Answer

Expert verified
a) The probability of A winning both her matches and B winning over C is .336. b) The probability of A winning both her matches is .56. c) The probability of A losing both her matches is .06. d) The probability that each person wins one match is .18.

Step by step solution

01

Determining the Probability of A Winning Both Her Matches & B Winning Over C

By multiplying the individual probabilities of independent events, the probability of A winning both her matches and B beats C is \(P(A_{bc} \cap B_c) = P(A_b) * P(A_c) * P(B_c) = .7 * .8 * .6 = .336\).
02

Calculating the Probability of A Winning Both Her Matches

Considering only the two matches involving A (against B and C), and since they are independent, the probability of A winning both is \(P(A_{bc}) = P(A_b) * P(A_c) = .7 * .8 = .56\).
03

Finding the Probability of A Losing Both Her Matches

The trick here is to find the probabilities of A losing to B and C. Since probability of A winning over B is .7, so probability of her losing is 1-.7 = .3. Similarly, probability of A losing to C is 1-.8 = .2. Multiply these to get the overall probability. So, \(P(A'_{bc}) = P(A'_b) * P(A'_c) = .3 * .2 = .06\).
04

Determining the Probability of Each Person Wins One Match

There are two scrambles where each person can win one match. These are: (A beats B, B beats C, C beats A) and (A beats C, C beats B, B beats A). Calculate each of these individually and add them. So, \(P(HV_1) = P(A_b) * P(B_c) * P(C_a) = .7 * .6 * (1-.8) = .084\), and \(P(HV_2) = P(A_c) * P(C_b) * P(B_a) = .8 * (1-.6) * (1-.7) = .096\). Adding these gives \(P(HV) = P(HV_1) + P(HV_2) = .084 + .096 = .18\).

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Key Concepts

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

Understanding Round-Robin Tournament Probability
When it comes to round-robin tournaments, every participant plays every other participant exactly once. This structure is common in various sports and games. In probability, when analyzing a round-robin tournament, we often want to determine the likelihood of various outcomes, such as a certain player winning all their matches.

In the given exercise, we delve into such a scenario with three friends鈥擜, B, and C鈥攅ach playing the others. By assuming that the matches' outcomes are independent events, we can simplify the calculations. Independence means the result of one match does not influence the outcomes of the other matches. This property is crucial because it allows us to multiply the probabilities of the separate events to find the probability of them all occurring in conjunction.

Let's take an example scenario where we try to find the probability of friend A winning both matches. Since these are independent events, we calculate it by multiplying the probability of A winning against B by the probability of A winning against C. Hence, the approach to solving these types of problems depends deeply on the notion of independent events and the multiplication rule of probability.
Calculating Winning Probabilities in Tournaments
When you're tackling problems that involve calculating winning probabilities, it's essential to remember the basics of probability theory. The probability of an event is a measure of the likelihood that the event will occur, and ranges from 0 (the event will not occur) to 1 (the event will certainly occur).

To illustrate, let鈥檚 consider the exercise where we calculate the probability of A winning both her matches. We find this by multiplying the probabilities of A鈥檚 independent victories. This method applies generally to calculate the probability of multiple independent events occurring. It's also worth noting that if you're asked about the probability of an event not occurring, you'll use the complement rule. This rule states that the probability of an event not occurring is equal to one minus the probability of the event occurring.

In our example, to find the probability of A losing both matches, we subtract the chances of A winning from one and then multiply those. So, understanding how to manipulate these basic probability rules is key when calculating winning probabilities in tournaments.
Combinatorial Probability Scenarios in Tournaments
In more complex tournaments, you may encounter combinatorial probability scenarios. These scenarios ask you to consider all possible ways an event can occur and are a step up from merely calculating the probability of isolated events.

For instance, in the round-robin tournament featuring friends A, B, and C, there鈥檚 a question about the probability of each person winning one match. This is a classic combinatorial problem鈥攖here are different combinations in which each player can win one match, and we need to consider each combination individually. This often involves identifying all possible scenarios that satisfy the condition (in this case, each player winning one match), calculating the probability of each scenario, and then adding these probabilities together to get the total probability.

As we see from the exercise, the final step is not just identifying the right scenarios but also ensuring that you are correctly applying the multiplication rule for independent events and then using the addition rule for mutually exclusive outcomes to arrive at the final probability figure for the event.

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Most popular questions from this chapter

USA Today (March 15,2001\()\) introduced a measure of racial and ethnic diversity called the Diversity Index. The Diversity Index is supposed to approximate the probability that two randomly selected individuals are racially or ethnically different. The equation used to compute the Diversity Index after the 1990 census was $$ \begin{aligned} 1-&\left[P(W)^{2}+P(B)^{2}+P(A I)^{2}+P(A P I)^{2}\right] \\ & \cdot\left[P(H)^{2}+P(\operatorname{not} H)^{2}\right] \end{aligned} $$ where \(W\) is the outcome that a randomly selected individual is white, \(B\) is the outcome that a randomly selected individual is black, \(A I\) is the outcome that a randomly selected individual is American Indian, \(A P I\) is the outcome that a randomly selected individual is Asian or Pacific Islander, and \(H\) is the outcome that a randomly sclected individual is Hispanic. The explanation of this index stared that 1\. \(\left[P(W)^{2}+P(B)^{2}+P(A I)^{2}+P(A P I)^{2}\right]\) is the probability that two randomly selected individuals are the same race 2\. \(\left[P(H)^{2}+P(\text { not } H)^{2}\right]\) is the probability that two randomly selected individuals are cither both Hispanic or both not Hispanic 3\. The calculation of the Diversity Index treats Hispanic ethnicity as if it were independent of race. a. What additional assumption about race must be made to justify use of the addition rule in the computarion of \(\left[P(W)^{2}+P(B)^{2}+P(A l)^{2}+P(A P I)^{2}\right]\) as the probability that two randomly selected individuals are of the same race? b. Three different probability rules are used in the calculation of the Diversity Index: the Complement Rule, the Addition Rule, and the Multiplication Rule. Describe the way in which each is used.

The paper "Predictors of Complementary Therapy Use among Asthma Patients Results of a Primary Care Survey"" (Health and Social Care in the Community 12008 h \(155-164\) ) included the accompanying rable. The table summarizes the responses given by 1077 asthma patients to two questions: Question 1: Do conventional asthma medications usually help your asthma symptoms? Question 2: Do you use complementary therapies (such as herbs, acupuncture, aroma therapy) in the treatment of your asthma? \begin{tabular}{l|cc} & Doesn't Use Complementary Therapies & Does Use Complementary Theraples \\ \hline Comventional Medica- & 816 & 131 \\ \multicolumn{1}{c|} { tions Usually Help } \\ Conventional Medi- & & \\ cations Usually Do Not Help & 103 & 27 \\ \hline \end{tabular} From this information, we can estimate that the proportion who use complementary therapies is \(\frac{131+27}{1077}=\frac{158}{1077}=.147 .\) For those who report that conventional medications usually help, the proportion who use complementary therapies is \(\frac{131}{947}=.138\) and for those who report that conventional medications usually do not help, the proportion who use comple- mentary therapies is \(\frac{27}{130}=.208\) If one of these 1077 patients is selected at random, are the outcomes selected patient reports that conventional medications wually help and selected patient uses complementary therapies independent or dependent? Explain.

Is ultrasound a reliable method for determining the gender of an unborn baby? Consider the following data on 1000 births, which are consistent with summary values that appeared in the online journal of Statistics Education ("New Approaches to Learning Probability in the First Statistics Course" 120011 ): \begin{tabular}{l|cc} & Ultrasound Predicted Female & Utrasound Predicted Male \\ \hline Actual Gender Is Female & 432 & 48 \\ Actual Gender is Male & 130 & 390 \\ \hline \end{tabular} Do you think that a prediction that a baby is male and a prediction that a baby is female are equally reliable? Explain, using the information in the table to calculate estimates of any probabilities that are relevant to your conclusion.

An article in the New york Times (March 2, 1994) reported that people who suffer cardiac arrest in New York City have only a 1 in 100 chance of survival. Using probability notation, an equivalent statement would be \(P(\) survival \()=.01\) for people who suffer a cardiac arrest in New York City (The artide attributed this poor survival rate to factors common in large cities: traffic congestion and the difficulty of finding victims in large buildings. Similar studies in smaller cities showed higher survival rates.) a. Give a relative frequency interpretation of the given probability. b. The research that was the basis for the New York Times article was a study of 2329 consecutive cardiac arrests in New York City. To justify the " 1 in 100 chance of survival" statement, how many of the 2329 cardiac arrest sufferers do you think survived? Explain.

The report "TV Drama/Comedy Viewers and Health Information" (www.cdc.gov/healthmarketing) describes the results of a large survey involving approximately 3500 people that was conducted for the Centers for Disease Control. The sample was selected in a way that the Centers for Disease Control believed would result in a sample that was representative of adult Americans, One question on the survey asked respondents if they had learned something new about a health issue or disease from a TV show in the previous 6 months. Consider the following outcomes: \(L=\) outcome that a randomly selected adult American reports learning something new about a health issue or disease from a TV show in the previous 6 months and \(F=\) outcome that a randomly selected adult American is female Data from the survey were used to estimate the following probabilities: $$ P(L)=.58 \quad P(F)=.50 \quad P(L \text { and } F)=.31 $$ Are the outcomes \(L\) and \(F\) independent? Use probabilities to justify your answer.
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