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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) \(=.8, P(\mathrm{~B}\) 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 \(\mathrm{B}\) 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.)

Short Answer

Expert verified
a. The probability that A wins both her matches and that B beats C is .336. b. The probability that A wins both her matches is .56. c. The probability that A loses both her matches is .06. d. The probability that each person wins one match is .228.

Step by step solution

01

Part a: Find the Probability that A wins both her matches and that B beats C

Since these are independent events, the probability that all these events occur is simply the product of their individual probabilities. Therefore, the calculation is \(P(\mathrm{A}\, \text{beats}\, \mathrm{B}) * P(\mathrm{A}\, \text{beats}\, \mathrm{C}) * P(\mathrm{B}\, \text{beats}\, \mathrm{C}) = .7 * .8 * .6 = .336\)
02

Part b: Find the Probability that A wins both her matches

This is simply the product of the probabilities that A beats B and that A beats C, which is \(P(\mathrm{A}\, \text{beats}\, \mathrm{B}) * P(\mathrm{A}\, \text{beats}\, \mathrm{C}) = .7 * .8 = .56\)
03

Part c: Find the Probability that A loses both her matches

We know that the probabilities that A beats B and C are .7 and .8, respectively. Therefore, the probability that A loses to B is \(1 - .7 = .3\) and that A loses to C is \(1 - .8 = .2\). The probability that both these events occur is their product, which is \(.3 * .2 = .06\)
04

Part d: Find the Probability that each person wins one match

For each person to win one match, either B must beat A and C must beat B (with A beating C), or C must beat A and B must beat C (with A beating B). The probability of the first scenario is \(P(\mathrm{B}\, \text{beats}\, \mathrm{A}) * P(\mathrm{C}\, \text{beats}\, \mathrm{B}) * P(\mathrm{A}\, \text{beats}\, \mathrm{C}) = .3 * .6 * .8 = .144\). The probability of the second scenario is \(P(\mathrm{C}\, \text{beats}\, \mathrm{A}) * P(\mathrm{B}\, \text{beats}\, \mathrm{C}) * P(\mathrm{A}\, \text{beats}\, \mathrm{B}) = .2 * .6 * .7 = .084\). Adding these two gives the total probability of each person winning one match, which is \(.144 + .084 = .228\)

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Key Concepts

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

Independent Events Probability
When discussing probabilities, it's important to understand what is meant by independent events. Independent events are outcomes that have no effect on one another. For example, flipping a coin and rolling a die are independent events because the result of the coin flip does not influence the roll of the die, and vice versa.

In the setting of a round-robin tournament, when we talk about independent events, we mean that the result of one match does not affect the outcome of another match. Given the problem with friends A, B, and C, it is assumed that the matches are played independently; that is, the chances of A beating B do not change because of the match between A and C or B and C.

It is fundamental to understand that the probability of compound independent events—multiple independent outcomes happening simultaneously—is calculated by multiplying their individual probabilities. If we take matches as independent events, the probability that A wins against B and also wins against C is the product of those two independent probabilities. Therefore, for calculating such probabilities, recognizing the independence of events is key to determining the correct computational approach.
Probability Calculations
The calculation of probabilities is central to understanding round-robin tournaments, and it applies to a vast array of scenarios in probability theory. Starting with basic principles, there are a few fundamental rules. One is the complementary rule, which states that the probability of an event occurring is 1 minus the probability of it not occurring. For instance, if player A has a 0.7 chance of beating player B, consequently, A has a 0.3 chance of not beating B, because 1 - 0.7 = 0.3.

To tackle more complex probability calculations, often you start with these basic rules and combine them to find probabilities of a sequence of events. If you are asked the probability of multiple independent events occurring, like A winning against both B and C, you multiply the individual probabilities together. This product rule is crucial in scenarios where several outcomes must occur in conjunction.

Using these techniques, you can calculate the probability of any selection of outcomes in a round-robin tournament, such as someone winning all their games or each player winning one game. As we've seen in our exercise examples, once the probabilities of basic outcomes are established, the bigger picture starts to get clearer with systematic calculations tailored to the specific questions.
Round-Robin Tournament Outcomes
In any round-robin tournament, each participant plays every other participant once. The calculation of possible outcomes in such a tournament can become quite intricate depending on the number of participants and the potential results of each match. In this example, with just three friends—A, B, and C—the number of potential outcomes is limited, making the calculations more straightforward but the concepts no less significant.

To calculate the overall outcomes, you must consider all possible ways each match can end based on the probability of each player winning or losing. For instance, when evaluating the scenario where each player wins one match, you must think through all the possible sequences of wins and losses that can lead to that result. It's often helpful to draw a diagram or chart to visualize the different possibilities and ensure that you're considering every combination.

Finally, to improve understanding of these concepts it can be beneficial to engage with real-world examples or simulate tournaments. Tools like probability trees or simulations can offer a dynamic illustration, enhancing comprehension of the potentially complex outcomes found in round-robin tournaments. Additionally, discussing the various scenarios and working through each step methodically, as we've done in our previous steps, can significantly improve understanding of round-robin tournament outcomes and probability calculations in general.

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Most popular questions from this chapter

A certain company sends \(40 \%\) of its overnight mail parcels by means of express mail service \(A_{1}\). Of these parcels, \(2 \%\) arrive after the guaranteed delivery time (use \(L\) to denote the event late delivery). If a record of an overnight mailing is randomly selected from the company's files, what is the probability that the parcel went by means of \(A_{1}\) and was late?

Two different airlines have a flight from Los Angeles to New York that departs each weekday morning at a certain time. Let \(E\) denote the event that the first airline's flight is fully booked on a particular day, and let \(F\) denote the event that the second airline's flight is fully booked on that same day. Suppose that \(P(E)=.7\), \(P(F)=.6\), and \(P(E \cap F)=.54 .\) a. Calculate \(P(E \mid F)\), the probability that the first airline's flight is fully booked given that the second airline's flight is fully booked. b. Calculate \(P(F \mid E)\).

The manager of a music store has kept records of the number of CDs bought in a single transaction by customers who make a purchase at the store. The accompanying table gives six possible outcomes and the estimated probability associated with each of these outcomes for the chance experiment that consists of observing the number of CDs purchased by the next customer at the store. $$ \begin{aligned} &\begin{array}{l} \text { Number of CDs } \\ \text { purchased } \end{array} & 1 & 2 & 3 & 4 & 5 & 6 \text { or more } \\ &\begin{array}{c} \text { Estimated } \\ \text { probability } \end{array} & .45 & .25 & .10 & .10 & .07 & .03 \end{aligned} $$ a. What is the estimated probability that the next customer purchases three or fewer CDs? b. What is the estimated probability that the next customer purchases at most three CDs? How does this compare to the probability computed in Part (a)? c. What is the estimated probability that the next customer purchases five or more CDs? d. What is the estimated probability that the next customer purchases one or two CDs? e. What is the estimated probability that the next customer purchases more than two CDs? Show two different ways to compute this probability that use the probability rules of this section.

Consider the chance experiment in which both tennis racket head size and grip size are noted for a randomly selected customer at a particular store. The six possible outcomes (simple events) and their probabilities are displayed in the following table: $$ \begin{array}{llll} && {\text { Grip Size }} \\ \text { Head size } & 4_{8}^{\frac{3}{8}} \text { in. } & 4 \frac{1}{2} \text { in. } & 4 \frac{5}{8} \text { in. } \\ \hline \text { Midsize } & O_{1}(.10) & O_{2}(.20) & O_{3}(.15) \\ \text { Oversize } & O_{4}(.20) & O_{5}(.15) & O_{6}(.20) \\ \hline \end{array} $$ a. The probability that grip size is \(4 \frac{1}{2}\) inches (event \(A\) ) is \(P(A)=P\left(O_{2}\right.\) or \(\left.O_{5}\right)=.20+.15=.35\). How would you interpret this probability? b. Use the result of Part (a) to calculate the probability that grip size is not \(4 \frac{1}{2}\) inches c. What is the probability that the racket purchased has an oversize head (event \(B\) ), and how would you interpret this probability? d. What is the probability that grip size is at least \(4 \frac{1}{2}\) inches?

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 article attributed this poor survival rate to factors common in large cities: traffic congestion and the difficulty of finding victims in large buildings.) 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.

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