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Chapter 6: Problem 54

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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Most popular questions from this chapter

The article "SUVs Score Low in New Federal Rollover Ratings" (San Luis Obispo Tribune, January 6,2001 ) gave information on death rates for various kinds of accidents by vehicle type for accidents reported to the police. Suppose that we randomly select an accident reported to the police and consider the following events: \(R=\) event that the selected accident is a single-vehicle rollover, \(F=\) event that the selected accident is a frontal collision, and \(D=\) event that the selected accident results in a death. Information in the article indicates that the following probability estimates are reasonable: \(P(R)=.06, P(F)=.60\), \(P(R \mid D)=.30, P(F \mid D)=.54\).

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