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Evaluate these binomial probabilities: a. \(C_{0}^{8}(.2)^{0}(.8)^{8}\) b. \(C_{1}^{8}(.2)^{1}(.8)^{7}\) c. \(C_{2}^{8}(.2)^{2}(.8)^{6}\) d. \(P(x \leq 1)\) when \(n=8, p=.2\) P(two or fewer successes)

Short Answer

Expert verified
Answer: The probability of having two or fewer successes is approximately 0.8053.

Step by step solution

01

a. Evaluate \(C_{0}^{8}(.2)^{0}(.8)^{8}\)

To evaluate the given expression, we need to find the combination (n choose k) and then raise the probabilities to the respective powers: \(C_{0}^{8} = \frac{8!}{0!(8-0)!} = 1\) Now, substitute the values into the given expression: \(C_{0}^{8}(.2)^{0}(.8)^{8} = 1 \cdot (1) \cdot (0.8)^{8} \approx 0.16777\)
02

b. Evaluate \(C_{1}^{8}(.2)^{1}(.8)^{7}\)

Calculate the combination (n choose k): \(C_{1}^{8} = \frac{8!}{1!(8-1)!} = 8\) Plug the values into the given expression: \(C_{1}^{8}(.2)^{1}(.8)^{7} = 8 \cdot (0.2) \cdot (0.8)^{7} \approx 0.33554\)
03

c. Evaluate \(C_{2}^{8}(.2)^{2}(.8)^{6}\)

Calculate the combination (n choose k): \(C_{2}^{8} = \frac{8!}{2!(8-2)!} = 28\) Plug the values into the given expression: \(C_{2}^{8}(.2)^{2}(.8)^{6} = 28 \cdot (0.2)^{2} \cdot (0.8)^{6} \approx 0.30199\)
04

d. Calculate \(P(x \leq 1)\) when \(n=8, p=.2\)

\(P(x \leq 1)\) means the probability of having 0 or 1 success in 8 trials when the probability of success is 0.2. So, we need to calculate \(P(x=0) + P(x=1)\): - We already calculated \(P(x=0)\) in part (a): \(\approx 0.16777\) - We already calculated \(P(x=1)\) in part (b): \(\approx 0.33554\) Now, add both probabilities: \(P(x \leq 1) = P(x=0) + P(x=1) \approx 0.16777 + 0.33554 = 0.50331\)
05

P(two or fewer successes)

To find the probability of having two or fewer successes in 8 trials when the probability of success is 0.2, we need to calculate \(P(x \leq 2)\): \(P(x \leq 2) = P(x=0) + P(x=1) + P(x=2)\) - We calculated \(P(x=0)\) in part (a): \(\approx 0.16777\) - We calculated \(P(x=1)\) in part (b): \(\approx 0.33554\) - We calculated \(P(x=2)\) in part (c): \(\approx 0.30199\) Now, add all probabilities: \(P(x \leq 2) = P(x=0) + P(x=1) + P(x=2) \approx 0.16777 + 0.33554 + 0.30199 = 0.8053\) Therefore, the probability of having two or fewer successes is approximately 0.8053.

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Key Concepts

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

Combination Formula
The Combination Formula is a crucial component in understanding binomial probabilities. It is used to determine how many different ways a particular number of successes can occur in a set number of trials. The formula is represented as \( C_{k}^{n} = \frac{n!}{k!(n-k)!} \). Here, \( n \) is the total number of trials, \( k \) is the number of successful trials, and the exclamation point \(!\) denotes factorial, which is the product of all positive integers up to that number.
For example, in our exercise, when calculating \( C_{0}^{8} \), it simplifies to \( \frac{8!}{0!(8-0)!} = 1 \). This means that there is only one way to have zero successes in eight tries. Understanding combinations is key for exploring binomial distributions, as it lays the groundwork for evaluating the likelihood of various outcomes.
Probability of Success
In any binomial probability scenario, it is essential to define the probability of success for a single trial. This probability is often denoted by \( p \). It represents the likelihood of the event of interest occurring in a single trial.
In our scenario, \( p = 0.2 \) indicates a 20% chance of success on each individual trial. The complementary probability of failure is denoted by \( q \), where \( q = 1 - p \). For our problem, \( q = 0.8 \) represents an 80% chance of not achieving success in a single trial. These probabilities are fundamental in calculating individual binomial probabilities for different values of \( k \), the number of successful events we're interested in.
Binomial Distribution
A binomial distribution comes into play when you're dealing with experiments consisting of a fixed number of independent trials, each with the same probability of success. It maps out probabilities of different numbers of successes that can be observed in these trials.
The binomial distribution is characterized mathematically by its probability mass function (PMF):\[ P(X=k) = C_{k}^{n} \cdot p^{k} \cdot (1-p)^{n-k} \]
Where \( X \) is the random variable representing the number of successes, \( n \) is the total number of trials, \( k \) is the number of actual successes, \( p \) is the probability of success, and \( (1-p) \) is the probability of failure.
  • In this way, each outcome's probability in a binomial distribution depends on both the number of ways it can happen (combinations) and the probability of those events happening.
  • Hence, the terms in our setup, like \( C_{1}^{8} \cdot (0.2)^{1} \cdot (0.8)^{7} \), form parts of the distribution that together describe the likelihoods of different numbers of successes across trials.
Cumulative Probability
Cumulative probability is a concept that involves summing up probabilities to find the likelihood of at most a certain number of successes within given trials. It represents the probability that a random variable is less than or equal to a specified value.
In our problem, cumulative probability helps us determine \( P(x \leq 1) \) or \( P(x \leq 2) \). It combines the probabilities of observing all number of successes from 0 up to a certain threshold.
  • For instance, \( P(x \leq 1) \) includes the probability of 0 successes and 1 success calculated in steps \(a\) and \(b\) respectively, giving us \(0.50331\).
  • To find \( P(x \leq 2) \), we add \( P(x=0)\), \( P(x=1)\), and \( P(x=2)\), culminating in a more extensive coverage of possible successful outcomes within the trials.
This approach is very useful in statistical analysis when making decisions based on thresholds of acceptable success rates.

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Most popular questions from this chapter

A Snapshot in USA Today shows that \(60 \%\) of consumers say they have become more conservative spenders. \({ }^{12}\) When asked "What would you do first if you won \(\$ 1\) million tomorrow?" the answers had to do with somewhat conservative measures like "hire a financial advisor," or "pay off my credit card," or "pay off my mortgage.' Suppose a random sample of \(n=15\) consumers is selected and the number \(x\) of those who say they have become conservative spenders recorded. a. What is the probability that more than six consumers say they have become conservative spenders? b. What is the probability that fewer than five of those sampled have become conservative spenders? c. What is the probability that exactly nine of those sampled are now conservative spenders.

Find \(P(x \leq k)\) for each of the following cases: a. \(n=20, p=.05, k=2\) b. \(n=15, p=.7, k=8\) c. \(n=10, p=.9, k=9\)

The recession has caused many people to use their credit cards far less. In fact, in the United States, \(60 \%\) of consumers say they are committed to living with fewer credit cards. \({ }^{15} \mathrm{~A}\) sample of \(n=400\) consumers with credit cards are randomly selected. a. What is the average number of consumers in the sample who said they are committed to living with fewer credit cards? b. What is the standard deviation of the number in the sample who said they are committed to living with fewer credit cards? c. Within what range would you expect to find the number in the sample who said they are committed to living with fewer credit cards? d. If only 200 of the sample of consumers said they were committed to living with fewer credit cards, would you consider this unusual? Explain. What conclusion might you draw from this sample information?

Find the mean and standard deviation for a binomial distribution with \(n=100\) and these values of \(p:\) a. \(p=.01\) b. \(p=.9\) c. \(p=.3\) d. \(p=.7\) e. \(p=.5\)

A city commissioner claims that \(80 \%\) of all people in the city favor private garbage collection in contrast to collection by city employees. To check the \(80 \%\) claim, you randomly sample 25 people and find that \(x\), the number of people who support the commissioner's claim, is \(22 .\) a. What is the probability of observing at least 22 who support the commissioner's claim if, in fact, \(p=.8 ?\) b. What is the probability that \(x\) is exactly equal to \(22 ?\) c. Based on the results of part a, what would you conclude about the claim that \(80 \%\) of all people in the city favor private collection? Explain.

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