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A soccer team estimates that they will score on \(8 \%\) of the corner kicks. In next week's game, the team hopes to kick 15 corner kicks. What are the chances that they will score on 2 of those opportunities?

Short Answer

Expert verified
The probability that the soccer team will score on 2 of the 15 corner kicks is calculated to be approximately 0.283.

Step by step solution

01

Identify Binomial Parameters

First, it's important to identify the parameters of the binomial distribution. The number of trials \(n\) is 15 (the number of corner kicks), the number of 'successes' \(k\) we are interested in is 2 (scoring from a corner kick), and the probability of 'success' \(p\) on any given trial is 0.08.
02

Apply Binomial Formula

Next, apply the binomial formula to calculate the probability. The formula is \(P(X=k) = C(n, k) * p^k * (1-p)^{n-k}\), where \(P(X=k)\) is the probability of \(k\) 'successes' in \(n\) trials, \(C(n, k)\) is the 'combination' function representing the number of possible combinations of \(n\) items taken \(k\) at a time.
03

Calculate Probability

Plug \(n=15\), \(k=2\), and \(p=0.08\) into the formula to calculate the probability: \[P(X=2) = C(15, 2) * 0.08^2 * (1-0.08)^{15-2}\] Calculate \(C(15, 2)\) as \( \frac{15!}{2!(15-2)!} = 105\), then substitute into the formula and compute the final result.

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Key Concepts

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

Binomial Parameters
The binomial distribution is a statistical model that describes the number of successes in a fixed number of independent trials. The parameters that define a binomial distribution include the number of trials (n), the number of successes (k), and the probability of success on a single trial (p).

In the context of our soccer team exercise, the number of trials is 15, which represents the corner kicks during a match. The two desired successes reflect the instances the team scores from those corner kicks. The probability of success (p) on any given trial is the chance of scoring from a corner kick, quantified as 8% or 0.08 in a decimal form. Understanding these binomial parameters is crucial to apply the appropriate statistical formula to estimate the probability of scoring exactly two goals.
Probability Calculation
Calculating probabilities with a binomial distribution involves using a specific formula: \[P(X=k) = {C(n, k) \cdot p^k \cdot (1-p)^{n-k}}\]. This formula determines the likelihood of getting exactly k successes (in our case, scoring goals) out of n trials, with a success probability of p per trial.

Applying this to our scenario, our aim is to find out the probability of the soccer team scoring on 2 of their 15 corner kicks. We use the given success rate of 8% per corner kick and calculate the probability by substituting our parameters into the binomial formula. Comprehension of this step is vital as it lays down the process of arriving at the probability value using the binomial distribution.
Combinatorial Function
The combinatorial function in the binomial probability formula, denoted as \(C(n, k)\), or sometimes \({n \choose k}\), refers to the number of combinations or ways to choose k items from a set of n items without regard to order.

For calculating \(C(n, k)\), the formula used is: \[C(n, k) = \frac{n!}{k!(n-k)!}\]. Here, the exclamation mark signifies factorial, which is the product of an integer and all the integers below it, down to one. In the soccer example, we use the combinatorial function to figure out how many different ways the team can score 2 goals out of 15 corner kicks. This function is a central element of the binomial formula and knowing how to calculate it enables us to further delineate the probability of a particular outcome.
Factorial
The factorial is a function that plays a significant role in both combinatorics and probability calculations. It's denoted by an exclamation point and is defined for a positive integer n as the product of all positive integers less than or equal to n. Mathematically, it's given as: \[ n! = n \times (n-1) \times (n-2) \times \ldots \times 2 \times 1 \].

For instance, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). Factorials are used to calculate combinations in our probability formula. In the soccer problem, when we need to compute \(C(15, 2)\), we are essentially using factorials to determine how many unique pairs of goals can be scored. Understanding factorials is essential for working with combinatorial functions and, consequently, with binomial probability calculations.

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Most popular questions from this chapter

A Department of Transportation report about air travel found that, nationwide, \(76 \%\) of all flights are on time. Suppose you are at the airport and your flight is one of 50 scheduled to take off in the next two hours. Can you consider these departures to be Bernoulli trials? Explain.

We have looked at various facts about hurricanes in previous chapters. Suppose we find that the arrivals of hurricanes can be modeled by a Poisson distribution with mean 2.45 . a. What's the probability of no hurricanes next year? b. What's the probability that during the next two years, there's exactly 1 hurricane?

Do these situations involve Bernoulli trials? Explain. a. You are rolling 5 dice and need to get at least two 6 's to win the game. b. We record the distribution of eye colors found in a group of 500 people. c. A manufacturer recalls a doll because about \(3 \%\) have buttons that are not properly attached. Customers return 37 of these dolls to the local toy store. Is the manufacturer likely to find any dangerous buttons?

About \(8 \%\) of males are color-blind. A researcher needs some color-blind subjects for an experiment and begins checking potential subjects. a. On average, how many men should the researcher expect to check to find one who is color-blind? b. What's the probability that she won't find anyone colorblind among the first 4 men she checks? c. What's the probability that the first color-blind man found will be the sixth person checked? d. What's the probability that she finds someone who is color-blind before checking the 10th man?

The basketball player in Exercise 55 has new sneakers, which he thinks improve his game. Over his past 40 shots, he's made 32 -much better than the \(55 \%\) he usually shoots. Do you think his chances of making a shot really increased? In other words, is making at least 32 of 40 shots really unusual for him? (Do you think it's his sneakers?)

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