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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 \(P(\) 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 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 B beats C is .336. b) The probability that A wins both her matches without considering who B beats is .56. c) The probability that A loses both her matches is .06. d) The probability that each person wins one match is .18.

Step by step solution

01

Probability that A wins both matches and B beats C

Since the outcomes are independent, the probability of multiple events happening is the product of their individual probabilities. So, the probability that A wins both her matches and B beating C is calculated as: \(P(\) A wins against B \(\cap \) A wins against C \(\cap \) B beats C\) = \(P(\) A beats B \(\) x \(P(\) A beats C \(\) x \(P(\) B beats C\)= .7 x .8 x .6 = .336.
02

Probability that A wins both her matches

The probability that A wins both her matches is calculated as: \(P(\) A wins against B \(\cap \) A wins against C\) = \(P(\) A beats B \(\) x \(P(\) A beats C \(\) = .7 x .8 = .56.
03

Probability that A loses against both her opponents

The probability that A loses both her games is 1 minus the probability that she wins. Therefore, she would lose to B with a probability of 1 - .7 = .3 and to C with a probability of 1 - .8 = .2. Using the same method as the other probabilities, \(P(\) A loses against B \(\cap \) A loses against C\) = \(P(\) A loses to B \(\) x \(P(\) A loses to C \(\)= .3 x .2 = .06.
04

Probability that each person wins one match

This can happen in two ways: A beats B, B beats C and C beats A, or A beats C, C beats B and B beats A. We calculate these two probabilities and add them together. From the first scenario: \(P(\) A beats B \(\) x \(P(\) B beats C \(\) x \(P(\) C beats A \(\) = .7 x .6 x .2 = .084. From the second scenario: \(P(\) A beats C \(\) x \(P(\) C beats B \(\) x \(P(\) B beats A \(\) = .8 x .4 x .3 = .096. Therefore, the total probability is .084 + .096 = .18.

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Key Concepts

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

Independent Events
In the realm of probability, understanding independent events is crucial. Independent events are situations in which the occurrence of one event does not affect the occurrence of another. Simply put, each event occurs separately from others.
Consider a round-robin tournament involving three friends: A, B, and C. When we say that the outcomes of their matches are independent, it implies that the result of one match does not influence the result of another. For instance, A beating B does not impact whether C can beat A or B can beat C.
A simple way to think about independent events is to imagine rolling dice. The outcome of one roll never affects the next. Similarly in our tournament, each match occurs independently. This makes calculations simpler, as we can directly multiply probabilities without considering any overlapping influences.
Multiplication Rule in Probability
The multiplication rule in probability is a handy tool, especially when dealing with independent events. It states that for independent events, the probability of multiple events occurring together is the product of their individual probabilities.
In the round-robin tournament, if you want to know the probability of A winning both matches and B beating C, you can use this rule. You simply multiply the individual probabilities:
  • Probability A beats B = 0.7
  • Probability A beats C = 0.8
  • Probability B beats C = 0.6
The combined probability is thus calculated as \( 0.7 \times 0.8 \times 0.6 = 0.336 \).
This rule is particularly powerful in scenarios where multiple independent outcomes are involved. It provides a straightforward way to derive complex probabilities from simpler ones.
Round-robin Tournament
A round-robin tournament is a competition format where each participant plays against every other participant. It's a fantastic way to ensure that all competitors face each other equally.
In our example involving friends A, B, and C, each will play against the others. This leads to a total of 3 matches: A vs. B, A vs. C, and B vs. C.
Round-robin tournaments stand out for their fairness. Every competitor has the same experience, and strategic calculations come into play. These tournaments provide thrilling probabilities to analyze, especially when outcomes are independent as in our scenario.
Conditional Probability
Conditional probability deals with the likelihood of an event occurring given that another event has already happened. While independent events don't rely on other occurrences, conditional probabilities depend on prior outcomes.
In a round-robin setup, conditional probabilities would become relevant if, for instance, one player's performance affects another's outcomes. However, in our scenario, the results are independent, so we do not need to apply conditional probability.
Understanding the contrast between independent and conditional probabilities can help in analyzing problems where some outcomes can affect others. In real-life tournaments, especially, performance in earlier matches might impact later ones, leading to conditional events.

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