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Consider two binomial experiments. a. The first binomial experiment consists of six trials. How many outcomes have exactly one success, and what are these outcomes? b. The second binomial experiment consists of 20 trials. How many outcomes have exactly 10 successes? exactly 15 successes? exactly 5 successes?

Short Answer

Expert verified
For the first experiment, there are 6 outcomes with exactly one success. In the second experiment, there are 184756 outcomes with exactly 10 successes, 15504 outcomes with exactly 15 successes, and 15504 outcomes with exactly 5 successes.

Step by step solution

01

Determine outcomes for the first experiment

For the first experiment, there are six trials and the problem asks to find outcomes having exactly one success. Use the formula for combination \[ C(n,k) = \frac{n!}{k!(n-k)!} \]. When n is 6 and k is 1, it becomes \( C(6,1) = 6 \), which means there are 6 outcomes that represent exactly one success.
02

Determine outcomes for the second experiment

For the second experiment, again use the combination formula to calculate outcomes with exactly 10, 15 and 5 successes, while there are total 20 trials.For 10 successes, it's \( C(20,10) = 184756 \).For 15 successes, it's \( C(20,15) = 15504 \).And finally, for 5 successes, it's \( C(20,5) = 15504 \). Each calculation uses the formula \[ C(n,k) = \frac{n!}{k!(n-k)!} \] with n as 20 and varying values of k.

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Key Concepts

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

Combinations
Combinations are an essential concept when dealing with problems where the arrangement order does not matter. This is particularly important in binomial experiments, where you might be interested in finding how many ways you can choose a subset of items from a larger set. For example, in a binomial experiment with multiple trials, we calculate the number of ways of obtaining a certain number of successes in a fixed number of trials.

To compute combinations, we use the formula for combination:
  • \(C(n,k) = \frac{n!}{k!(n-k)!}\)
Here, \( n \) is the total number of items, \( k \) is the number of items to choose, and \(!\) represents a factorial. The factorial of a number \( n \) is the product of all positive integers up to \( n \).

This formula can be applied to various situations in probability, aiding in calculating outcomes with specific properties, such as exactly one success, which was demonstrated in the original exercise.
Binomial Coefficient
The binomial coefficient is directly related to combinations and is often visualized in the context of binomial expansions. In practice, the binomial coefficient tells us the number of ways to choose \( k \) successes out of \( n \) trials in a binomial experiment. It is denoted by \( C(n,k) \), and its calculation involves using the same formula we use for combinations.
Let's further illustrate this: given an experiment with 20 trials, to find how many outcomes have exactly 10 successes, you compute \(C(20,10) = \frac{20!}{10! \cdot (20-10)!} = 184756\).
The binomial coefficient supports various areas of combinatorics beyond just calculating probabilities in binomial experiments, making it a fundamental element of discrete mathematics.
Probability Theory
Probability Theory is the mathematical framework that allows us to analyze and predict random events, a cornerstone for understanding binomial experiments. It involves calculating the likelihood of various outcomes, providing a structured approach to uncertainty. In binomial experiments, each trial is independent and has only two possible outcomes: success or failure.
In these experiments, probabilities are calculated using both the concept of combinations and the binomial distribution formula, which considers the success probability \( p \) and the number of trials \( n \). This formula helps in determining the probability of a specific number of successes out of \( n \) trials:
  • \(P(X=k) = C(n,k) \cdot p^k \cdot (1-p)^{n-k}\)
This equation captures the beauty of probability theory by quantifying how likely it is to achieve \( k \) successes, considering the inherent randomness of the process.
Understanding these concepts is crucial not only for solving textbook problems but also for comprehending real-world random processes.

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Most popular questions from this chapter

A machine that cuts corks for wine bottles operates in such a way that the distribution of the diameter of the corks produced is well approximated by a normal distribution with mean \(3 \mathrm{~cm}\) and standard deviation \(0.1 \mathrm{~cm}\). The specifications call for corks with diameters between \(2.9\) and \(3.1 \mathrm{~cm}\). A cork not meeting the specifications is considered defective. (A cork that is too small leaks and causes the wine to deteriorate; a cork that is too large doesn't fit in the bottle.) What proportion of corks produced by this machine are defective?

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A gas station sells gasoline at the following prices (in cents per gallon, depending on the type of gas and service): \(315.9,318.9,329.9,339.9,344.9\), and 359.7. Let \(y\) denote the price per gallon paid by a randomly selected customer. a. Is \(y\) a discrete random variable? Explain. b. Suppose that the probability distribution of \(y\) is as follows: $$ \begin{array}{lrrrrrr} y & 315.9 & 318.9 & 329.9 & 339.9 & 344.9 & 359.7 \\ p(y) & .36 & .24 & .10 & .16 & .08 & .06 \end{array} $$ What is the probability that a randomly selected customer has paid more than $$\$ 3,20$$ per gallon? Less than $$\$ 3.40$$ per gallon? c. Refer to Part (b), and calculate the mean value and standard deviation of \(y .\) Interpret these values.

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