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Chapter 8: Problem 31

According to a study, \({ }^{13}\) the probability that a randomly selected teenager watched a rented video at least once during a week was \(.71 .\) What is the probability that at least 8 teenagers in a group of 10 watched a rented movie at least once last week?

Short Answer

Expert verified
The probability that at least 8 teenagers in a group of 10 watched a rented movie at least once last week is: \[P(X \geq 8) = P(X=8) + P(X=9) + P(X=10)\] Using the binomial formula, we get: \[P(X \geq 8) = C(10, 8) \cdot 0.71^8 \cdot (1-0.71)^{10-8} + C(10, 9) \cdot 0.71^9 \cdot (1-0.71)^{10-9} + C(10, 10) \cdot 0.71^{10} \cdot (1-0.71)^{10-10}\] Calculating these values, we find that the probability is approximately 0.4847.

Step by step solution

01

Identify the given values

In this problem, we are given: - Probability of success (p) = 0.71 - Number of trials (n) = 10 - Number of successes (at least 8)
02

Calculate the probability for each case

We need to find the probability that at least 8 teenagers in a group of 10 watched a rented movie at least once last week. This means we can have 8, 9, or 10 successes. We will calculate the probability for each case and then sum them up. For each case, we will use the binomial probability formula: \[P(X=k) = C(n, k) \cdot p^k \cdot (1-p)^{n-k}\] Where: - \(P(X=k)\) is the probability of having k successes in n trials - \(C(n, k)\) is the number of combinations of choosing k successes from n trials - p is the probability of success - k is the number of successes, and - n is the total number of trials
03

Calculate the probabilities for 8, 9, and 10 successes

Now, we will calculate the probabilities for 8, 9, and 10 successes using the binomial probability formula: 1. For 8 successes: \[P(X=8) = C(10, 8) \cdot 0.71^8 \cdot (1-0.71)^{10-8}\] 2. For 9 successes: \[P(X=9) = C(10, 9) \cdot 0.71^9 \cdot (1-0.71)^{10-9}\] 3. For 10 successes: \[P(X=10) = C(10, 10) \cdot 0.71^{10} \cdot (1-0.71)^{10-10}\]
04

Calculate the sum of probabilities for at least 8 successes

Finally, to find the probability that at least 8 teenagers in a group of 10 watched a rented movie at least once last week, we will sum up the probabilities for 8, 9, and 10 successes: \[P(X \geq 8) = P(X=8) + P(X=9) + P(X=10)\] By plugging in the values calculated in step 3, we can find the probability: \[P(X \geq 8) = P(X=8) + P(X=9) + P(X=10)\]
05

Calculate the final probability

Once you have summed up the probabilities, you will get the final probability of at least 8 teenagers in a group of 10 watching a rented movie at least once last week.

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Key Concepts

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

Probability of Success
Understanding the probability of success is crucial when working with binomial probability models. It refers to the likelihood of a particular outcome you're hoping for in one trial. In our exercise, we see the probability of success, denoted as 'p', is 0.71. This means that, according to a study, there's a 71% chance that a randomly selected teenager watched a rented video at least once during a week.

This probability is an estimate based on historical data or given hypotheses and is essential for any further calculations in the binomial probability distribution. To make informed predictions about future outcomes, knowing the exact value of 'p' is necessary. When the probability of success is high, as in our case, we anticipate that the outcome will occur frequently, which will significantly influence the total probability when multiple trials are considered.
Number of Trials
The number of trials, often symbolized by 'n', is a count of how many times an experiment is conducted or how many opportunities there are for an event to occur. In the binomial setting, each trial is assumed to be independent, meaning the outcome of one trial doesn't influence the others, and the probability remains consistent across all trials.

In the provided example, we are considering a group of 10 teenagers, which gives us a total of 10 trials. If we're to predict occurrences of an event within this group, it's vital to count all the trials because the more trials we have, the more opportunities there are for the event (in this case, teenagers watching a rented video) to occur. A larger number of trials can also provide a more reliable overall probability, but every trial is still constrained by the fixed probability of success 'p'.
Probability Formula
The binomial probability formula is the engine behind determining the likelihood of a certain number of successes across a number of trials. The formula is expressed as:
\[P(X=k) = C(n, k) \cdot p^k \cdot (1-p)^{n-k}\]
where 'P(X=k)' refers to the probability of exactly 'k' successes out of 'n' trials, 'C(n, k)' is the combinations function which calculates how many different ways 'k' successes can occur during 'n' trials, 'p' is the probability of success per trial, and '(1-p)' is the probability of failure.

By utilizing this formula, we can find out the probabilities of different counts of successes. For example, if we want to determine the likelihood of at least 8 teenagers watching a rented movie in our scenario with 10 trials, we calculate the probability of exactly 8, exactly 9, and exactly 10 teenagers watching, and sum these probabilities up. This sum represents the collective probability of at least 8 of the events occurring, taking into account all the different ways that outcome can be achieved.

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Most popular questions from this chapter

Calculate the expected value of the given random variable \(X .\) [Exercises \(23,24,27\), and 28 assume familiarity with counting arguments and probability (Section 7.4).] \(\nabla 30\) darts are thrown at a dartboard. The probability of hitting a bull's-eye is \(\frac{1}{5}\). Let \(X\) be the number of bull's-eyes hit.

The probability of a plane crashing on a single trip in 1990 was .00000087. Find the approximate probability that in \(100,000,000\) flights, there will be more than 110 crashes.

Exercises \(45-50\) are based on the following information, gathered from student testing of a statistical software package called MODSTAT. \(^{56}\) Students were asked to complete certain tasks using the software, without any instructions. The results were as follows. (Assume that the time for each task is normally distributed. Find the probability that a student will take at least 10 minutes to complete Task 3 .

The Acme Insurance Company is launching a drive to generate greater profits, and it decides to insure racetrack drivers against wrecking their cars. The company's research shows that, on average, a racetrack driver races four times a year and has a 1 in 10 chance of wrecking a vehicle, worth an average of \(\$ 100,000\), in every race. The annual premium is \(\$ 5,000\), and Acme automatically drops any driver who is involved in an accident (after paying for a new car), but does not refund the premium. How much profit (or loss) can the company expect to earn from a typical driver in a year? HINT [Use a tree diagram to compute the probabilities of the various outcomes.]

Video Arcades Your company, Sonic Video, Inc., has conducted research that shows the following probability distribution, where \(X\) is the number of video arcades in a randomly chosen city with more than 500,000 inhabitants: \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|} \hline\(x\) & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline\(P(X=x)\) & \(.07\) & \(.09\) & \(.35\) & \(.25\) & \(.15\) & \(.03\) & \(.02\) & \(.02\) & \(.01\) & \(.01\) \\ \hline \end{tabular} a. Compute the mean, variance, and standard deviation (accurate to one decimal place). b. As CEO of Startrooper Video Unlimited, you wish to install a chain of video arcades in Sleepy City, U.S.A. The city council regulations require that the number of arcades be within the range shared by at least \(75 \%\) of all cities. What is this range? What is the largest number of video arcades you should install so as to comply with this regulation?
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