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Chapter 3: Problem 35

The probability Coach Sears' basketball team wins any given game is \(0.8\), regardless of any prior win or loss. If her team plays five games, what is the probability it wins more games than it loses?

Short Answer

Expert verified
The probability that Coach Sears' team wins more games than it loses, given that the probability of winning any given game is 0.8, is obtained by adding the binomial probabilities of winning 3, 4 or 5 games out of 5 played.

Step by step solution

01

Understand the Scenario and Identify the Parameters

For Coach Sears' basketball team, the probability of winning every single game is 0.8. The team is going to play five games. The outcomes are independent. We basically need to compute the probabilities of the team winning 3, 4, or all of the 5 games.
02

Calculation

We have to calculate the probability of winning different outcomes. We use the binomial distribution formula which is as follows: \(P(X=k) = C(n,k) \cdot p^k \cdot (1-p)^{n-k}\) where \(n\) is the total number of trials (or games in our case), \(p\) is the probability of winning a trial (or game), \(X=k\) is the event that there are \(k\) wins, and \(C(n,k)\) is the combination of \(n\) items taken \(k\) at a time. Here, \(C(n,k) = n! / [k!(n-k)!]\) where '!' denotes factorial. Now, calculate the different outcomes: Probability of winning 3 games: \(P(X=3) = C(5,3) \cdot 0.8^3 \cdot (1-0.8)^{5-3}\) Probability of winning 4 games: \(P(X=4) = C(5,4) \cdot 0.8^4 \cdot (1-0.8)^{5-4}\) Probability of winning 5 games: \(P(X=5) = C(5,5) \cdot 0.8^5 \cdot (1-0.8)^{5-5}\)
03

Summing Up The Probabilities

We have to add up the probabilities of winning 3, 4 and 5 games because these are mutually exclusive events, meaning that only one of these events can occur at a time. Hence, the overall probability of winning more games than losing is given by: \(P = P(X=3) + P(X=4) + P(X=5)\)

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Key Concepts

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

Probability
Probability is a fundamental concept in statistics and mathematics that measures how likely an event is to occur. It ranges from 0 to 1, where 0 indicates impossibility and 1 indicates certainty. In the context of Coach Sears' basketball team, we're interested in the probability of winning a game, which is 0.8. This means, on average, the team is very likely to win any single game they play.

When solving problems like this, we often deal with random events. Each game played by the team can be seen as a separate trial, with two possible outcomes: "win" or "lose". Understanding this helps in computing the probability of different scenarios, such as winning a certain number of games out of several played. Probability calculations guide us in predicting and understanding the likelihood of such outcomes.
Binomial Formula
The binomial formula is a powerful and widely-used tool in probability for determining the likelihood of a given number of "successes" in a set number of trials, each with the same probability of success. It is especially useful when dealing with scenarios where the outcomes are binary (e.g., win/lose).

The formula itself is:
  • \[ P(X=k) = C(n,k) \cdot p^k \cdot (1-p)^{n-k} \]
where:
  • \( n \) is the total number of trials (games).
  • \( p \) is the probability of success on a single trial (probability of winning a game, 0.8 in this case).
  • \( X=k \) represents the event of achieving \( k \) successes (wins).
  • \( C(n,k) \) is the number of ways to choose \( k \) successes in \( n \) trials, calculated as \( n! / [k!(n-k)!] \).
In Coach Sears' scenario, to find out the probability of winning more games than losing, it is necessary to calculate probabilities for winning 3, 4, or 5 games, and sum them up. Each of these probabilities can be easily computed using the binomial formula.
Independent Events
Independent events are those whose outcomes do not influence each other. In probability theory, this means the outcome of one event does not change the likelihood of another event occurring. For Coach Sears鈥� basketball team, the assumption that each game's outcome is independent is key.

This independence means the probability of winning each game stays the same regardless of the result of the previous games. Thus, if the team wins one game, it doesn't affect their odds of winning the next one, keeping the win probability at a constant 0.8 for every game.

Understanding independence is crucial when using the binomial distribution, as it underpins the math behind calculating the probability of multiple wins. Without independence, the calculations and the probabilities derived from the binomial formula would not be valid.

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Most popular questions from this chapter

An integer is selected at random from 3 through 17 inclusive. If \(A\) is the event that a number divisible by 3 is chosen and \(B\) is the event that the number exceeds 10 , determine \(\operatorname{Pr}(A), \operatorname{Pr}(B), \operatorname{Pr}(A \cap B)\), and \(\operatorname{Pr}(A \cup B)\). How is \(\operatorname{Pr}(A \cup B)\) related to \(\operatorname{Pr}(A), \operatorname{Pr}(B)\), and \(\operatorname{Pr}(A \cap B) ?\)

Alice tosses a fair coin seven times. Find the probability she gets four heads given that (a) her first toss is a head; (b) her first and last tosses are heads.

Let \(\mathscr{S}\) be the sample space for an experiment \(\mathscr{E}\) and let \(A, B\) be events from \(\mathscr{P}\), where \(\operatorname{Pr}(A)=0.4, \operatorname{Pr}(B)=\) \(0.3\), and \(\operatorname{Pr}(A \cap B)=0.2\). Determine \(\operatorname{Pr}(\bar{A}), \operatorname{Pr}(\bar{B})\), \(\operatorname{Pr}(A \cup B), \operatorname{Pr}(\overline{A \cup B}), \operatorname{Pr}(A \cap \bar{B}), \operatorname{Pr}(\bar{A} \cap B)\) \(\operatorname{Pr}(A \cup \bar{B})\), and \(\operatorname{Pr}(\bar{A} \cup B) .\)

In alpha testing a new software package, a software engineer finds that the number of defects per 100 lines of code is a random variable \(X\) with probability distribution: $$ \begin{array}{c|cccc} \boldsymbol{x} & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} \\ \hline \boldsymbol{P r}(\boldsymbol{X}=\boldsymbol{x}) & 0.4 & 0.3 & 0.2 & 0.1 \end{array} $$ Find (a) \(\operatorname{Pr}(X>1)\); (b) \(\operatorname{Pr}(X=3 \mid X \geq 2)\); (c) \(E(X)\); and (d) \(\operatorname{Var}(X)\).

The exit door at the end of a hallway is open half of the time. On a table by the entrance to this hallway is a box containing 10 keys, but only one of these keys opens the exit door at the end of the hallway. Upon entering the hallway Marlo selects two of the keys from the box. What is the probability she will be able to leave the hallway via the exit door, without returning to the box for more keys?
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