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Chapter 5: Problem 117

Two teams, \(\mathrm{A}\) and \(\mathrm{B}\), will play a best-of-seven series, which will end as soon as one of the teams wins four games. Thus, the series may end in four, five, six, or seven games. Assume that each team has an equal chance of winning each game and that all games are independent of one another. Find the following probabilities. a. Team A wins the series in four games. b. Team A wins the series in five games. c. Seven games are required for a team to win the series.

Short Answer

Expert verified
a) The probability that Team A wins the series in four games is 0.0625. b) The probability that Team A wins the series in five games is \(4 \cdot P(4; 5, 0.5).\) c) The probability that seven games are required for a team to win the series is \(P(3; 6, 0.5) \cdot 1.\)

Step by step solution

01

Team A wins the series in four games

To win the series in four games, Team A must win all four games. Given the games are independent and each team has an equal chance of winning, we can use the binomial probability formula to compute this. Here \(P=0.5\), \(n=4\), \(x=4\) So, \(P(4; 4, 0.5) = C(4, 4) \cdot (0.5)^4 \cdot (0.5)^{4 -4}= 1 \cdot (0.5)^4 = 0.0625.\)
02

Team A wins the series in five games

To win the series in five games, Team A must win four out of the five games, with the loss occurring anywhere in the first four games. This gives us four scenarios where Team A could win (i.e., the loss happens in game 1, 2, 3, or 4). Each of these cases should have a similar probability, so we calculate the probability for one case and then multiply by four.Here \(P=0.5\), \(n=5\), \(x=4\) So, \(P(4; 5, 0.5) = C(5, 4) \cdot (0.5)^4 \cdot (0.5)^{5 -4}.\)This gives the probability of one scenario, and there are four such scenarios, so multiply this by four. Therefore, the probability that Team A wins the series in five games is \(4 \cdot P(4; 5, 0.5).\)
03

Seven games are required for a team to win the series

Seven games are required for a team to win if each team wins three of the first six games. The 7th game can then be won by either team A or B, hence we consider both possibilities. First calculate the probability of a team winning 3 out of 6 games. Here \(P=0.5\), \(n=6\), \(x=3\) So, \(P(3; 6, 0.5) = C(6, 3) \cdot (0.5)^3 \cdot (0.5)^{6 -3}.\)This is the case for the first 6 games. And for the last game, either team could win, so multiply this probability by \(2 \cdot (0.5)=1\). Therefore, the probability that 7 games are required to win the series is \(P(3; 6, 0.5) \cdot 1.\)

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Key Concepts

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

Probability Theory
Probability Theory is the branch of mathematics concerned with analyzing random events. If an outcome is unpredictable yet has a pattern when repeated numerous times, it falls under probability. Probability helps us quantify the likelihood of various outcomes. This is achieved by identifying all possible outcomes of an event and assigning each a value between 0 and 1. Here, 0 indicates impossibility, while 1 signifies certainty. Understanding probabilities allows us to make informed predictions about uncertain events, like predicting the outcome of a sporting series or a coin flip. For example, in the context of the series between Team A and Team B, each team has a 50% chance, or a probability of 0.5, to win any given match. This doesn't mean that one will win exactly half the time over a few games, but over many plays, these probabilities will reveal the overall pattern.
Independent Events
In probability theory, two events are considered independent if the occurrence of one does not affect the other. This concept is crucial when dealing with sequences of events, such as a series of game matches or coin flips. The independence of events ensures that each game between Team A and Team B has no influence from past games. Hence, the outcome of one game is unaffected by previous games. In mathematical terms, two events A and B are independent if \(P(A \cap B) = P(A) \cdot P(B)\). For example, if it rains, it doesn鈥檛 affect your chance of flipping heads on a coin, thus they remain independent.This concept allows us to analyze each game separately, treating each as a distinct event. It simplifies calculations as the overall probability becomes a product of individual probabilities.
Binary Outcomes
Binary outcomes refer to results that can only have two possible values. These are common in scenarios where there is a clear distinction between two possible and opposite results, such as win/lose or yes/no. In the game series between Team A and Team B, each match ends with one of two outcomes: Team A wins or Team B wins. This binary nature allows us to employ the binomial probability formula. Binomial probability is useful when analyzing situations with a fixed number of trials, independent events, and constant probability of success or failure. Utilizing binary outcomes, we can assess the likelihood of Team A winning 4 games out of 5. Each game holds a constant success probability (winning), which guides our probability assessments effectively. This method is key in projecting the series outcome over a set number of games.

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Most popular questions from this chapter

An average of \(6.3\) robberies occur per day in a large city. a. Using the Poisson formula, find the probability that on a given day exactly 3 robberies will occur in this city. b. Using the appropriate probabilities table from Appendix \(C\), find the probability that on a given day the number of robberies that will occur in this city is i. at least 12 ii. at most 3 iii. 2 to \(\overline{6}\)

A professional basketball player makes \(85 \%\) of the free throws he tries. Assuming this percentage will hold true for future attempts, find the probability that in the next eight tries, the number of free throws he will make is a. exactly 8 b. exactly 5

Let \(x\) be the number of cars that a randomly selected auto mechanic repairs on a given day. The following table lists the probability distribution of \(x\). $$ \begin{array}{l|ccccc} \hline x & 2 & 3 & 4 & 5 & 6 \\ \hline P(x) & .05 & .22 & .40 & 23 & .10 \\ \hline \end{array} $$ Find the mean and standard deviation of \(x\), Give a brief interpretation of the value of the mean.

Although Borok's Electronics Company has no openings, it still receives an average of \(3.2\) unsolicited applications per week from people seeking jobs. a. Using the Poisson formula, find the probability that this company will receive no applications next week. b. Let \(x\) denote the number of applications this company will receive during a given week. Using the Poisson probabilities table from Appendix \(\mathrm{C}\), write the probability distribution table of \(x\) c. Find the mean, variance, and standard deviation of the probability distribution developed in part b.

Let \(N=16, r=10\), and \(n=5 .\) Using the hypergeometric probability distribution formula, find a. \(P(x=5)\) b. \(P(x=0) \quad\) c. \(P(x \leq 1)\)
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