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Chapter 8: Problem 23

The probability that a randomly chosen person in the Netherlands connects to the Internet immediately upon waking is approximately \(.25 .^{10}\) What is the probability that, in a randomly selected sample of five people, two connect to the Internet immediately upon waking? HINT [See Example 2.]

Short Answer

Expert verified
The probability that in a randomly selected sample of five people, two of them connect to the internet immediately upon waking is approximately 0.2627 or 26.27%.

Step by step solution

01

Calculate the number of combinations of choosing 2 successes from 5 trials

We will use the formula for combinations: \[C(n, k) = \frac{n!}{k!(n-k)!}\] where \(n!=n\times(n-1)\times(n-2)\times\cdots\times1\). In our case, \(n=5\) and \(k=2\): \[C(5, 2) = \frac{5!}{2!(5-2)!}=\frac{5\times4\times3\times2\times1}{(2\times1)(3\times2\times1)} = \frac{120}{(2)(6)} = 10\]
02

Calculate p^k and (1-p)^(n-k)

In our case, \(p=0.25\), \(k=2\), \(n=5\), and \(1-p=0.75\): \[p^k = (0.25)^2 = 0.0625\] \[(1-p)^{(n-k)}=(0.75)^{(5-2)}=(0.75)^3=0.421875\]
03

Calculate the binomial probability

Now, we can calculate the binomial probability using the formula: \[P(X=k) = C(n,k)p^k(1-p)^{(n-k)}\] In our case: \[P(X=2) = 10\times0.0625\times0.421875 = 0.262695\] So, the probability that in a randomly selected sample of five people, two of them connect to the internet immediately upon waking is approximately 0.2627 or 26.27%.

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Key Concepts

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

Understanding Binomial Probability
Binomial probability is a fundamental concept in probability theory. It helps us calculate the likelihood of a specific number of successes in a fixed number of independent trials of a binary event. Each trial results in one of two outcomes - success or failure.
For example, consider five people, and we want to know the probability that two out of these five connect to the Internet immediately upon waking.
The formula for calculating binomial probability is:
  • \( P(X=k) = C(n, k) p^k (1-p)^{n-k} \)
Here, \( C(n, k) \) represents the number of combinations, \( p \) symbolizes the probability of success in one trial, and \( 1-p \) indicates the probability of failure. Five trials, choosing two successes, and each has a 0.25 probability makes this approach methodical.
Exploring Combinatorics
Combinatorics is the branch of mathematics that deals with combinations and permutations. In the context of binomial probability, we're specifically interested in how we can select a certain number of successes from the total trials.
This is calculated using the combination formula:
  • \( C(n, k) = \frac{n!}{k!(n-k)!} \)
Where the factorial \( n! \) represents the product of all positive integers up to \( n \).
In our case, we have 5 trials (n=5) and we want to select 2 successes (k=2). Solving \( C(5, 2) \) provided us with 10, which signifies that there are 10 unique ways to have two people connect out of five.
Probability Distributions Demystified
Probability distributions describe how the probability of various outcomes is distributed for a random variable. A binomial distribution specifically models the number of successes in a given number of trials for a binary outcome experiment.
It's defined by two parameters:
  • \( n \): Number of trials
  • \( p \): Probability of a success on an individual trial
For any random variable \( X \) that follows a binomial distribution, the probability of obtaining exactly \( k \) successes is given by the binomial probability formula. This distribution is discrete, meaning it describes non-continuous outcomes , like our Internet connection scenario.
Additionally, since the probabilities must sum to 1, the distribution can help in calculating probabilities for different numbers of successes in the trials effectively.

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Most popular questions from this chapter

A roulette wheel has the numbers 1 through 36,0 , and \(00 .\) Half of the numbers from 1 through 36 are red, and a bet on red pays even money (that is, if you win, you will get back your \(\$ 1\) plus another \(\$ 1\) ). How much do you expect to win with a \(\$ 1\) bet on red? HINT [See Example 4.]

Calculate the expected value, the variance, and the standard deviation of the given random variable \(X .\) You calculated the expected values in the last exercise set. Round all answers to two decimal places.) \(X\) is the number of tails that come up when a coin is tossed three times.

I \(\vee\) Supermarkets A survey of supermarkets in the United States yielded the following relative frequency table, where \(X\) is the number of checkout lanes at a randomly chosen supermarket: \(^{49}\) \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|} \hline \(\boldsymbol{x}\) & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline \(\boldsymbol{P}(\boldsymbol{X}=\boldsymbol{x})\) & \(.01\) & \(.04\) & \(.04\) & \(.08\) & \(.10\) & \(.15\) & \(.25\) & \(.20\) & \(.08\) & \(.05\) \\ \hline \end{tabular} a. Compute the mean, variance, and standard deviation (accurate to one decimal place). b. As financial planning manager at Express Lane Mart, you wish to install a number of checkout lanes that is in the range of at least \(75 \%\) of all supermarkets. What is this range according to Chebyshev's inequality? What is the least number of checkout lanes you should install so as to fall within this range?

The following figures show the price of silver per ounce, in dollars, for the 10 -business day period Feb. 2 -Feb. 13 , $$ \begin{aligned} &2009: 22 \\ &12.4,12.4,12.4,12.8,12.9,13.0,13.0,13.4,13.3,13.4 \end{aligned} $$ Find the sample mean, median, and mode(s). What do your answers tell you about the price of silver?

Following is a sample of the day-byday change, rounded to the nearest 100 points, in the Dow Jones Industrial Average during 10 successive business days around the start of the financial crisis in October \(2008:^{20}\) $$ -100,400,-200,-500,200,-300,-200,900,-100,200 $$ Compute the mean and median of the given sample. Fill in the blank: There were as many days with a change in the Dow above \(\quad\) points as there were with changes below that.
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