Problem 7 - Algebra 2 and Trigonometry

Question

In \(7-14,\) answers can be rounded to four decimal places. The probability that our team will win a basketball game is \(\frac{2}{3} .\) What is the probability that they will win exactly 5 of the next 7 games?

Step-by-Step Solution

Answer

The probability that the team will win exactly 5 of the next 7 games is approximately 0.2704.

Step 1: Identify the Probability of Winning a Single Game

The probability given for the team winning a single game is \( \frac{2}{3} \). This value will be used as \( p \) in further calculations.

Step 2: Identify the Number of Trials

The problem involves the next 7 games, so the number of trials \( n \) is 7. We are calculating the probability of winning exactly 5 out of these 7 games.

Step 3: Identify the Number of Desired Wins

We want to find the probability of winning exactly 5 games, so the number of desired successful outcomes \( k \) is 5.

Step 4: Calculate the Probability of Losing a Game

The probability of losing a game is \( q = 1 - p = 1 - \frac{2}{3} = \frac{1}{3} \).

Step 5: Use the Binomial Probability Formula

The binomial probability formula is given by: \[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]Substitute the known values into this formula: \[ P(X = 5) = \binom{7}{5} \left( \frac{2}{3} \right)^5 \left( \frac{1}{3} \right)^{7-5} \]

Step 6: Calculate the Binomial Coefficient

Compute the binomial coefficient \( \binom{7}{5} = \frac{7!}{5!2!} = 21 \).

Step 7: Calculate the Probability

Substitute the values into the expression: \[ P(X = 5) = 21 \left( \frac{32}{243} \right) = 21 \times \frac{32}{729} \]Perform the calculations: \[ P(X = 5) = \frac{672}{729} \approx 0.2704 \]Thus, the probability that the team will win exactly 5 of the next 7 games is approximately 0.2704.

Key Concepts

Probability

Probability is a fundamental concept in statistics and is all about measuring the likelihood of an event happening. It is often represented as a number between 0 and 1, where 0 indicates the event will certainly not occur, and 1 indicates the event will certainly occur.
For example, if you have a probability of \(\frac{2}{3}\) for your team winning a game, it means that if the game were played many times under the same conditions, the team would win about two out of every three times. In the case of calculating probabilities, having this basic understanding is essential.

• A probability of 0.5 means the event has a 50% chance of occurring.
• A probability closer to 0 suggests a lesser likelihood.
• A probability closer to 1 suggests a higher likelihood.

Understanding probability helps in making predictions and decisions about the future, as it was crucial in determining the team's odds of winning exactly 5 out of 7 games in this scenario.

Binomial Coefficient

The binomial coefficient is an essential part of calculating probabilities in a binomial distribution. It represents the number of ways you can choose \( k \) successful outcomes from \( n \) trials. In mathematical terms, it's expressed using the formula \( \binom{n}{k} \).
This can be computed through factorials, articulated as \( \frac{n!}{k!(n-k)!} \). Here, "factorial" means you multiply a series of descending natural numbers. For example, \(5!\) is 5 multiplied by 4, then 3, and so forth, which equals 120.
In the exercise, the binomial coefficient \( \binom{7}{5} \) calculates the number of ways to choose 5 wins out of 7 games. It simplifies to \( \frac{7!}{5!2!} \), which equals 21.

• The binomial coefficient is crucial in determining combinations in probability scenarios.
• It assists in creating a path to calculate actual probabilities in binomial distributions.

Knowing how to compute these coefficients effectively allows one to solve much more complex problems involving probability distribution.

Binomial Distribution

Binomial distribution is a probability distribution that models the number of successful outcomes in a fixed number of independent trials, each with the same probability of success. In simpler terms, it evaluates how likely it is to achieve a set number of successes in a series of attempts, given the pursuit and its success probability.
In our case study, this distribution helped in calculating the probability of the basketball team winning exactly 5 out of 7 games. Binomial distribution uses the formula:

• \( P(X = k) = \binom{n}{k} p^k q^{n-k} \)

Here:

• \( n \) is the number of trials,
• \( k \) is the number of successful outcomes desired,
• \( p \) is the probability of success on a single trial,
• \( q = 1-p \) is the probability of a failure on a single trial.

By plugging in \( p = \frac{2}{3} \) and \( q = \frac{1}{3} \), and using the binomial coefficient calculated earlier (21), we determined the desired probability (around \(0.2704\)). This approach is incredibly useful in scenarios with a predictable sequence of events with two possible outcomes, such as win or lose.